$$\eqalign{ & {t_{\frac{1}{2}}} = 3.8\,{\text{day}} \cr & \therefore \lambda = \frac{{0.693}}{{{t_{\frac{1}{2}}}}} = \frac{{0.693}}{{3.8}} = 0.182 \cr} $$
If the initial number of atom is $$a = {A_0}$$  then after time $$t$$ the number of atoms is $$\frac{a}{{20}} = A.$$  We have to find $$t.$$
$$\eqalign{ & t = \frac{{2.303}}{\lambda }\log \frac{{{A_0}}}{A} = \frac{{2.303}}{{0.182}}\log \frac{a}{{\frac{a}{{20}}}} \cr & = \frac{{2.303}}{{0.182}}\log 20 = 16.46\,{\text{days}}\left( {{\text{approx}}} \right) \cr} $$

$$\eqalign{ & {N_1} = {N_0}{e^{ - {\lambda _1}t}} = {N_0}{e^{ - \frac{t}{\tau }}}\,......\left( {\text{i}} \right) \cr & {\text{as}}\,\tau = \frac{1}{{{\lambda _1}}} \cr & {N_2} = {N_0}{e^{ - {\lambda _2}t}} = {N_0}{e^{ - \frac{t}{{5\tau }}}}\,......\left( {\text{ii}} \right)\,{\text{as }}5\tau = \frac{1}{{{\lambda _2}}} \cr} $$
Adding (i) and (ii) we get
$$N = {N_1} + {N_2} = {N_0}\left( {{e^{ - \frac{t}{\tau }}} + {e^{ - \frac{t}{{5\tau }}}}} \right)$$
(A) is NOT the correct option as there is a time $$\tau $$ for which $$N$$ is constant which means for time $$\tau $$ there is no process of radioactivity which does not makes sense. (B) and (C) shows intermediate increase in the number of radioactive atom which is IMPOSSIBLE as $$N$$ will only decrease exponentially.