Answer
$E = 1.52\times 10^{-8}~N/C$
Work Step by Step
We can find the magnitude of the electric field $E$:
$E = \frac{1}{4~\pi~\epsilon_0}~\frac{\vert q \vert}{r^2}$
$E = \frac{1}{(4~\pi)~(8.854\times 10^{-12}~F/m)}~\frac{4.23\times 10^{-15}~C}{(50~m)^2}$
$E = 1.52\times 10^{-8}~N/C$
