Answer
$V=0.562mV$
Work Step by Step
$V=\frac{Kq}{d}+\frac{Kq}{d}+\frac{K(-q)}{d}+\frac{K(-q)}{2d}$
$V=\frac{Kq}{d}(1+1-1-\frac{1}{2})$
$V=\frac{Kq}{d}(\frac{1}{2})$
We plug in the known values to obtain:
$V=\frac{9\times 10^9\times 5\times 10^{-15}}{0.04}(\frac{1}{2})$
$V=0562\times 10^{-3}=0.562mV$