Explanation

(b) Let $R\ and\ r$ be the radii of bigger and each smaller drop. Charge remains conserved.
Hence, charge on bigger drop
$= 27 \times$charge on smaller drop
$ie,$ $q^{'} = 27q$
Now, before and after coaleseing, volume remains same.
That is,
$\frac{4}{3}\pi R^{3} = 27 \times \frac{4}{3}\pi r^{3}$
$\therefore\ R = 3r$
Hence, capacitance of bigger drop
$C^{'} = 4\pi\varepsilon_{0}R = 4\pi\varepsilon_{0}(3r)$
$= 3\left( 4\pi\varepsilon_{0}r \right) = 3C$