Explanation

(c) 64 drops have formed a signal drop of radius $R$. Volume of big drop = Volume of small’ drops
$\therefore\frac{4}{3}\pi R^{3} = 64 \times \frac{4}{3}\ \pi r^{3}$
$\Rightarrow \ \ \ \ \ \ R\ = 4r$
So, the total current is
$\ \ \ \ \ Q_{total} = 64q\ \ $
As, $C^{'} = \frac{Q}{V}and\ V = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{R};C^{'} = 4\pi\varepsilon_{0}R$
$C^{'} = \left( 4\pi\varepsilon_{0} \right)4r$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{'} = \ 4C$