In coalescing into a single drop charge remains conserved. Also volume before and after coalescing remains same.
Let $R$ and $r$ be the radii of bigger and each smaller drop. Charge remains conserved.
Hence, charge on bigger drop
$= 27 × $charge on smaller drop
i.e., $q' = 27\,q$
Now, before and after coalescing, 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}\left(3r\right) $
$= 3 \left(4\pi\varepsilon_{0}r\right) $
$= 3C$