If n drops, each of capacitance C, coalesce to form a single big drop, the capacitance of the big drop will be

If p is the density of a small drop and r its radius, then the mass of each small drop is m$=\frac{4\mathrm{\pi }}{3}{r}^3\rho$. If n such drops coalesce to form a big drop of radius R, then the mass of the big drop is nm$=\frac{4\mathrm{\pi }}{3}{R}^3\rho$. Hence $R=n^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}r$

Now, the capacitance of a sphere is proportional to its radius. Hence the capacitance of the big drop will be $C\text{'}=n^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}C$