Answer
$V=13$ kV
Work Step by Step
We can find electric potential as:
$V=\frac{Q_1}{4\pi \epsilon_{\circ}d}+\frac{3Q_1}{8\pi \epsilon_{\circ}d}-\frac{8Q_1}{16\pi \epsilon_{\circ}d}$
$V=\frac{Q_1}{8\pi \epsilon_{\circ}d}$
We plug in the known values to obtain:
$V=\frac{(8.99\times 10^9)(30\times 10^{-9})}{2(0.01)}=1.3\times 10^4=13\times 10^3=13$ kV
