Materials like black velvet or lamp black come close to ideal black bodies, but the best practical realization of an ideal black body is a small hole leading into a cavity maintained at constant temperature as this absorbs $$98\% $$  of the radiation incident on them. Cavity approximating an ideal black body is shown in the figure. Radiation entering the cavity has little chance of leaving before it is completely absorbed.

We know that
$$\eqalign{ & {\lambda _m}T = {\text{Constant}} \cr & {\lambda _A} < {\lambda _B} < {\lambda _C} \cr & {\text{So, }}{T_A} > {T_B} > {T_C} \cr & \left\{ {\because \,\,{T_A} = \frac{C}{{3 \times {{10}^{ - 7}}}},{T_B} = \frac{C}{{4 \times {{10}^{ - 7}}}},{T_C} = \frac{C}{{5 \times {{10}^{ - 7}}}}} \right\} \cr & Q = e\sigma A{T^4} \cr & e = 1\,\,{\text{black }}{\text{body}} \cr & \therefore \,\,Q = \sigma A{T^4} \cr & \therefore \,\,{Q_A} = \sigma .\pi {\left( {2 \times {{10}^{ - 2}}} \right)^2} \times \frac{{{C^4}}}{{27 \times {{10}^{ - 28}}}} \cr & {Q_B} = \sigma .\pi {\left( {4 \times {{10}^{ - 2}}} \right)^2} \times \frac{{{C^2}}}{{64 \times {{10}^{ - 28}}}} \cr & {\text{and, }}{Q_C} = \sigma .\pi {\left( {6 \times {{10}^{ - 2}}} \right)^2} \times \frac{{{C^2}}}{{625 \times {{10}^{ - 28}}}} \cr} $$
From comparison $${Q_B}$$ is maximum.

Heat transfer of glass bulb from filament is through radiation. A medium is required for convection process.
As a bulb is almost evacuated, heat from the filament is transmitted through radiation.

Black board paint is quite approximately equal to black bodies.

According to Newton's law of cooling, the rate of loss of heat of a body is directly proportional to the difference in temperatures of the body and the surroundings, provided the difference in temperature is small, not more than $${30^ \circ }C.$$
∴ Average rate of fall of temperature $$ \propto $$ average temperature excess
i.e., $$\frac{{dT}}{{dt}} \propto \left( {{T_t} - {T_s}} \right)$$
$$ \Rightarrow \frac{{dT}}{{dt}} = K\left( {{T_t} - {T_s}} \right)$$
According to question for 1^st and 2^nd case,
$$\eqalign{ & \frac{{50.1 - 49.9}}{5} = k\left[ {\frac{{50.1 + 49.9}}{2} - 30} \right]\,......\left( {\text{i}} \right) \cr & \frac{{40.1 - 39.9}}{{t'}} = k\left[ {\frac{{40.1 + 39.9}}{2} - 30} \right]\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing Eq. (i) by Eq. (ii), we get
$$\frac{2}{5} \times \frac{{t'}}{2} = \frac{{20}}{{10}} \Rightarrow t' = 10\,s$$

According to Stefan's law,
$$\eqalign{ & E \propto {T^4} \cr & \propto {\left( {t + 273} \right)^4}K\,\,\left[ {{{727}^ \circ }C = \left( {727 + 273} \right)K} \right] \cr & \propto {\left( {727 + 273} \right)^4}K \cr & \propto {\left( {1000} \right)^4}K \cr} $$

According to Stefan-Boltzmann law, amount of heat energy $$\left( E \right)$$  radiated per second by unit area of a body is directly proportional to the fourth power of absolute temperature $$\left( T \right)$$  of the body
i.e. $$E \propto {T^4}\,\,{\text{or}}\,\,E = \sigma {T^4}$$
If $$T$$ is doubled, $$E$$ becomes $${\left( 2 \right)^4}$$ times (i.e. 16 times).

The energy radiated per second by a black body is given by Stefan's Law
$$\frac{E}{t} = \sigma {T^4} \times A,$$   where $$A$$ is the surface area.
$$\frac{E}{t} = \sigma {T^4} \times 4\,\pi {r^2}$$    ($$\because $$ For a sphere, $$A = 4\pi {r^2}$$   )
Case (i) : $$\frac{E}{t} = 450,T = 500\,K,r = 0.12\,m$$
$$\therefore \,\,450 = 4\,\pi \sigma {\left( {500} \right)^4}{\left( {0.12} \right)^2}\,\,\,\,.....\left( {\text{i}} \right)$$
Case (ii) : $$\frac{E}{t} = \,? ,T = 1000\,K,r = 0.06\,m$$
$$\therefore \,\,\frac{E}{t} = 4\,\pi \sigma {\left( {1000} \right)^4}{\left( {0.06} \right)^2}\,\,\,.....\left( {{\text{ii}}} \right)$$
Dividing (ii) and (i), we get
$$\eqalign{ & \frac{{\frac{E}{t}}}{{450}} = \frac{{{{\left( {1000} \right)}^4}{{\left( {0.6} \right)}^2}}}{{{{\left( {500} \right)}^4}{{\left( {0.12} \right)}^2}}} \cr & = \frac{{{2^4}}}{{{2^2}}} \cr & = 4 \cr & \Rightarrow \,\,\frac{E}{t} = 450 \times 4 \cr & = 1800\,W \cr} $$

$$\eqalign{ & \frac{{50 - 49.9}}{5} = K\left( {\frac{{50 + 49.9}}{2} - 30} \right)\,......\left( {\text{i}} \right) \cr & \frac{{40 - 39.9}}{t} = K\left[ {\frac{{40 + 39.9}}{2} - 30} \right]\,......\left( {{\text{ii}}} \right) \cr} $$
From equations (i) and (ii), we get $$t \approx 10\,s.$$

$$Q = \sigma At\left( {{T^4} - T_0^4} \right)$$
If $$T,{T_0},\sigma $$   and $$t$$ are same for both bodies, then
$$\frac{{{Q_{{\text{sphere}}}}}}{{{Q_{{\text{cube}}}}}} = \frac{{{A_{{\text{sphere}}}}}}{{{A_{{\text{cube}}}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}}\,......\left( {\text{i}} \right)$$
But according to problem,
Volume of sphere = Volume of cube
$$ \Rightarrow \frac{4}{3}\pi {r^3} = {a^3} \Rightarrow a = r{\left( {\frac{4}{3}\pi } \right)^{\frac{1}{3}}}$$
Substituting the value of $$a$$ in equation (i), we get
$$\frac{{{Q_{{\text{sphere}}}}}}{{{Q_{{\text{cube}}}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}} = \frac{{4\pi {r^2}}}{{6{{\left\{ {{{\left( {\frac{4}{3}\pi } \right)}^{\frac{1}{3}}}r} \right\}}^2}}} = {\left( {\frac{\pi }{6}} \right)^{\frac{1}{3}}}:1$$