Answer
$476\;\rm V$
Work Step by Step
We know that the electric potential of a charged sphere on its surface is given by
$$V =\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R}\tag 1$$
So we need to find the charge and the radius of the final drop [the merge of two drops].
The charge is conserved, so the final charge of the final drop is
$$Q=Q_1+Q_2=0.1 +0.1$$
$$Q=\bf 0.2\;\rm nC\tag 2$$
We know that the two mercury drops have the same potential on their surfaces. We also know that
$$V_1=V_2=300$$
$$\dfrac{1}{4\pi \epsilon_0}\dfrac{Q_1}{R_1}=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q_2}{R_2}$$
and since $Q_1=Q_2$, so $R_1=R_2$
Hence,
$$\dfrac{1}{4\pi \epsilon_0}\dfrac{Q_1}{R_1}=V_1$$
$$R_1=R_2=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q_1}{V_1}\tag 3$$
Now we need to find the radius of the final drop, the volume of the final drops is given by
$${\rm V}_{net}={\rm V}_1+{\rm V}_2=\dfrac{4\pi R^3}{3}$$
$$ \dfrac{4\pi R_1^3}{3}+\dfrac{4\pi R_1^3}{3}=\dfrac{4\pi R^3}{3}$$
Hence,
$$R^3=2R_1^3$$
$$R=\sqrt[3]{2} R_1$$
Plug from (3),
$$R=\dfrac{1}{4\pi \epsilon_0}\dfrac{\sqrt[3]{2} Q_1}{V_1} $$
Plug the known;
$$R=(9\times 10^9)\dfrac{\sqrt[3]{2} (0.1\times 10^{-9})}{300} $$
$$R=\bf 3.78\times 10^{-3}\;\rm m\tag 4$$
Plug from (4) and (2) into (1),
$$V =\dfrac{1}{4\pi \epsilon_0}\dfrac{0.2\times 10^{-9}}{3.78\times 10^{-3}} $$
$$V =(9\times 10^9)\dfrac{0.2\times 10^{-9}}{3.78\times 10^{-3}} $$
$$V=\color{red}{\bf 476}\;\rm V$$
