Given $$M = 8 \times {10^{22}}\,A{m^2}$$
$$d = {R_e} = 6.4 \times {10^6}m$$
Earth’s magnetic field, $$B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}}$$
$$\eqalign{ & = \frac{{4\pi \times {{10}^{ - 7}}}}{{4\pi }} \times \frac{{2 \times 8 \times {{10}^{22}}}}{{{{\left( {6.4 \times {{10}^6}} \right)}^3}}} \cr & \cong 0.6\,Gauss \cr} $$

The total magnetic moment per unit volume.
$$\eqalign{ & = \frac{{\left( {8.52 \times {{10}^{28}}} \right) \times \left( {2 \times 9.27 \times {{10}^{ - 24}}} \right)}}{1} \cr & = 1.58 \times {10^6}\,\left. A \right|m \cr} $$

Curi’s Law $${\chi _m} = \frac{{{\mu _0}c}}{T}$$

$${T_1} = \frac{{60}}{{10}} = 6s\,{\text{and}}\,{T_2} = \frac{{60}}{{15}} = 4s$$
For identical magnets,
$$\frac{{{M_1}}}{{{M_2}}} = \frac{{T_2^2}}{{T_1^2}} = \frac{{{4^2}}}{{{6^2}}} = \frac{{16}}{{36}} = \frac{4}{9}.$$

Magnetic moment is from $$S$$ to $$N$$
So, $${M_{{\text{net}}}} = \sqrt {{M^2} + {M^2} + 2{M^2}\cos \theta } $$
$${M_{{\text{net}}}}$$ will be maximum if $${\cos \theta }$$  is maximum. $${\cos \theta }$$  will be maximum when $$\theta $$ will be minimum.
So, at $$\theta = {30^ \circ },{M_{{\text{net}}}}$$    will be maximum.

$$\eqalign{ & {\delta _1} = {40^ \circ },{\delta _2} = {30^ \circ },\delta = ? \cr & \cot \delta = \sqrt {{{\cot }^2}{\delta _1} + {{\cot }^2}{\delta _2}} \cr & = \sqrt {{{\cot }^2}{{40}^ \circ } + {{\cot }^2}{{30}^ \circ }} \cr & \cot \delta = \sqrt {{{1.19}^2} + 3} = 2.1 \cr & \therefore \delta = {25^ \circ }\,\,{\text{i}}{\text{.e}}{\text{.}}\,\,\delta < {40^ \circ } \cr} $$

The volume of the cube is
$$V = {\left( {{{10}^6} - m} \right)^3} = {10^{ - 18}}{m^3}$$
Net dipole moment $${m_{{\text{net}}}}$$
$$\eqalign{ & = 8 \times {10^{10}} \times 9 \times {10^{ - 24}}A{m^2} \cr & = 72 \times {10^{ - 14}}A{m^2} \cr} $$
Intensity of magnetization is magnetic moment developed per unit volume.
∴ magnetization, $$M = \frac{{{m_{{\text{net}}}}}}{{{\text{volume}}}}$$
$$ = \frac{{72 \times {{10}^{ - 14}}A{m^2}}}{{{{10}^{ - 18}}{m^3}}} = 7.2 \times {10^5}A{m^{ - 1}}$$

We know that, magnetic dipole moment
$$M = niA\cos \theta \,{\text{i}}{\text{.e}}{\text{.,}}\,M \propto \cos \theta $$
When two magnetic fields are inclined at an angle of $${75^ \circ }$$ the equilibrium will be at $${30^ \circ },$$ so $$\cos \theta = \cos \left( {{{75}^ \circ } - {{30}^ \circ }} \right) = \cos {45^ \circ } = \frac{1}{{\sqrt 2 }}$$
$$\eqalign{ & \frac{x}{{\sqrt 2 }} = \frac{{15}}{2} \cr & \therefore x \approx 11 \cr} $$

$$\eqalign{ & \tan \theta = \frac{{{B_V}}}{{{B_H}}} = 1 \cr & \therefore \theta = {45^ \circ } \cr} $$

Soft iron $$ \to $$ lowre tentivity and steel $$ \to $$ high retentivity
so option (C) is correct.