Answer
$V=2.5\times 10^6V$
Work Step by Step
We know that:
$V=2\times \frac{Kq}{r}.....eq(1)$
Next, we find $r$:
$r=\sqrt{(\frac{d}{2})^2+(\frac{d}{2})^2}=\sqrt{(\frac{0.02}{2})^2+(\frac{0.02}{2})^2}=0.014142m$
We plug in the known values in eq(1) to obtain:
$V=2\times \frac{9\times 10^9(2\times 10^{-6})}{0.014142}=2.5\times 10^6V$