Answer
$V=500V$
Work Step by Step
If we add the ratio of each drop of water, we get $R_{T}=2R$. The charge will be $Q=2q$, let's apply the formula:
$$V=\frac{1}{4 \pi \epsilon_{0}} \frac{Q}{R_{T}}=\frac{1}{4 \pi \epsilon_{0}} \frac{2q}{2R}=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{R}=500V$$
spite of both drops add themselves, the potential is the same
