Answer
$V_A=6\times 10^4V$
Work Step by Step
We can find electric potential at corner $A$ as
$V_A=\frac{1}{4\pi \epsilon_{\circ}}(\frac{q_1}{b}+\frac{q_2}{a})$
We plug in the known values to obtain:
$V_A=\frac{1}{4\times 3.1416\times 8.85\times 10^{-12}}(\frac{-5\times 10^{-6}}{0.15}+\frac{2\times 10^{-6}}{0.05})$
$V_A=6\times 10^4V$