Answer
$E_{s,net} = 2.54\times 10^{-6}~N/C$
Work Step by Step
Note that the horizontal components of the electric field due to each electron cancels out, and the net electric field is the sum of the vertical components.
We can find the magnitude of $E_{s,net}$:
$E_{s,net} = 2\times \frac{1}{4~\pi~\epsilon_0}~\frac{\vert q \vert}{r^2}~sin~\theta$
$E_{s,net} = \frac{1}{2~\pi~\epsilon_0}~\frac{\vert q \vert}{(\sqrt{2}~R)^2}~sin~45^{\circ}$
$E_{s,net} = \frac{1}{(2~\pi)~(8.854\times 10^{-12}~F/m)}~\frac{1.6\times 10^{-19}~C}{[(\sqrt{2})(0.0200~m)]^2}~sin(45^{\circ})$
$E_{s,net} = 2.54\times 10^{-6}~N/C$
