64 identical mercury drops combine to form one bigger drop. The capacitance of the bigger drop, as compared to that of a smaller drop will be 

If  r  is radius of the bigger drop and r is radius of each of the smaller drops, then
$\frac{4}{3}\pi R^3=64\left(\frac{4}{3}\pi r^3\right)$
Or,  R = 4r.

Capacitance of spherical conductor $\propto$ radius.

Hence, the capacitance becomes 4 times.