512 identical drops of mercury are charged

$Q=512 q$

Volume $_{\mathrm{i}}=$ Volume $_{\mathrm{f}}$

$512 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$

$2^{9} r^{3}=R^{3}$

$\mathrm{R}=8 \mathrm{r}$

$2=\frac{\mathrm{kq}}{\mathrm{r}}$

$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{R}}=\frac{\mathrm{k} 512 \mathrm{q}}{8 \mathrm{r}}$

$\mathrm{V}=128$