Answer
$E = 1.1\times 10^5~N/C$
Work Step by Step
We can find the electric field strength 2.0 mm above the center of the electrode.
$E = \frac{\sigma}{2~\epsilon_0}$
$E = \frac{Q}{2~\epsilon_0~A}$
$E = \frac{8.0\times 10^{-8}~C}{(2)(8.85\times 10^{-12}~C^2/N~m^2)(0.20~m)(0.20~m)}$
$E = 1.1\times 10^5~N/C$
