Answer
$q=2.2\times 10^{-6}C$
Work Step by Step
We know that:
$E=\frac{1}{4\pi \epsilon_{\circ}}\frac{q}{r^2}$
or $E=K\frac{q}{r^2}$ (where $K=\frac{1}{4\pi \epsilon_{\circ}}$)
This can be rearranged as:
$q=\frac{Er^2}{K}$
We plug in the known values to obtain:
$q=\frac{5.0\times 10^7(0.020)^2}{8.99\times 10^9}=2.2\times 10^{-6}C$
