Explanation

(a) Let $r$ be radius of each small drop and $R$ be radius of bigger drop
As the volume remains constant
$\therefore\frac{4}{3}\pi R^{3} = n \times \frac{4}{3}\pi r^{3} \Rightarrow R = n^{1/3}r$
Capacitance of each small drop, $C_{0} = 4\pi\varepsilon_{0}r$
Charge of each small drop, $q_{0} = C_{0}V = 4\pi\varepsilon_{0}rV$
Charge of bigger drop, $Q = nq_{0}$
Capacitance of bigger drop, $C = 4\pi\varepsilon_{0}R$
Potential of bigger drop, $= \frac{Q}{C} = \frac{nq_{0}}{4\pi\varepsilon_{0}R}$
$= \frac{n(4\pi\varepsilon_{0}rV)}{4\pi\varepsilon_{0}R} = nV\left( \frac{r}{R} \right) = n\left( \frac{1}{n^{1/3}}\ \right)V = n^{2/3}V$