Answer
$3.5\ pC$
Work Step by Step
We know that
$E=\frac{Kq}{r^2}+\frac{Kq}{r^2}$
$\implies E=\frac{2Kq}{r^2}$
This can be rearranged as:
$q=\frac{r^2E}{2K}$
We plug in the known values to obtain:
$q=\frac{(\frac{1}{2}\times 0.075)^2(45)}{2(8.99\times 10^9)}=3.5\ pC$
