Explanation

(a) Let radius of big drop be $R$and of smaller be $r$. Then,
Volume of bigger drop $= 8 \times$volume of single drop$\ $
Also, since shape of drop is assumed spherical, volume of a sphere of radius $a\ is\frac{4}{3}\pi a^{3}$
$\therefore\frac{4}{3}\pi R^{3} = 8 \times \frac{4}{3}\pi r^{3}$
$\Rightarrow \ = 2r\ $
Capacitance of sphere of radius $a$ is
$\ \ \ \ C = 4\pi\varepsilon_{0}a$
$\therefore\ \ \ $Capacitance of big drop
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C = 4\pi\varepsilon_{0}R$
Capacitance of small drop
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{'} = \ 4\ \pi\varepsilon_{0}(2r)$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{C'}{C} = \ \frac{4\ \pi\varepsilon_{0}(2r)}{4\pi\varepsilon_{0}r} = \frac{2}{1}$