If eight charged same water drops each with a radius of 1mm and a charge of ${10}^{-10}C$  merge into a single drop. Then the potential of the big drop is

r be the radius of small drop

R be the radius of big drop

volume is conserved 

so 

$V_{big}=8 V_{small}$

$\frac{4}{3}\mathrm{\pi } {\mathrm{R}}^3 = 8\times \frac{4}{3}{\mathrm{\pi r}}^3$

$\Rightarrow \mathrm{R}= 2\mathrm{r}$

$$\begin{gathered} \mathrm{potential} \mathrm{of} \mathrm{big} \mathrm{drop} \mathrm{is} {\mathrm{V}}_{\mathrm{big}}= \frac{1}{4{\mathrm{\pi \epsilon }}_0} \frac{8\mathrm{q}}{\mathrm{r}} \\ \mathrm{put} \mathrm{q} \mathrm{and} \mathrm{r} \mathrm{values} \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{equation} \mathrm{we} \mathrm{get} \mathrm{V} = 3600\mathrm{v} \end{gathered}$$