Answer
$V=2.9KV$
Work Step by Step
We know that
$V=V_1+V_2$
$\implies V=\frac{Kq_1}{r_1}+\frac{Kq_2}{r_2}$
We plug in the known values to obtain:
$V=\frac{9\times 10^9\times 1\times 10^{-8}}{0.03}+\frac{9\times 10^9\times (-3\times 10^{-8})}{2.0-0.03}=2.9\times 10^3=2.9KV$
