Answer
$E = 3.46\times 10^7~N/C$
Work Step by Step
We can find the value of $z$ where the electric field is a maximum:
$E = \frac{Qz}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}}$
$\frac{dE}{dz} = \frac{(Q)[4\pi~\epsilon_0~(z^2+R^2)^{3/2}]-(Qz)(3/2)[4\pi~\epsilon_0~(z^2+R^2)^{1/2}](2z)}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}} = 0$
$(Q)[4\pi~\epsilon_0~(z^2+R^2)^{3/2}] = (Qz)(3/2)[4\pi~\epsilon_0~(z^2+R^2)^{1/2}](2z)$
$z^2+R^2 = 3z^2$
$2z^2 = R^2$
$z = \frac{R}{\sqrt{2}}$
We can find the maximum magnitude:
$E = \frac{Qz}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}}$
$E = \frac{Q\frac{R}{\sqrt{2}}}{4\pi~\epsilon_0~[(\frac{R}{\sqrt{2}})^2+R^2]^{3/2}}$
$E = \frac{Q~R}{4~\sqrt{2}~\pi~\epsilon_0~(3R^2/2)^{3/2}}$
$E = \frac{Q}{4~\sqrt{2}~\pi~\epsilon_0~(1.5)^{3/2}~R^2}$
$E = \frac{4.00\times 10^{-6}~C}{(4~\sqrt{2}~\pi)~(8.854\times 10^{-12}~F/m)~(1.5)^{3/2}~(0.0200~m)^2}$
$E = 3.46\times 10^7~N/C$
