Answer
$E$ is a maximum at a distance of $~~1.70~cm~~$ from the ring's center.
Work Step by Step
We can find the value of $z$ where the electric field is a maximum:
$E = \frac{Qz}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}}$
$\frac{dE}{dz} = \frac{(Q)[4\pi~\epsilon_0~(z^2+R^2)^{3/2}]-(Qz)(3/2)[4\pi~\epsilon_0~(z^2+R^2)^{1/2}](2z)}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}} = 0$
$(Q)[4\pi~\epsilon_0~(z^2+R^2)^{3/2}] = (Qz)(3/2)[4\pi~\epsilon_0~(z^2+R^2)^{1/2}](2z)$
$z^2+R^2 = 3z^2$
$2z^2 = R^2$
$z = \frac{R}{\sqrt{2}}$
$z = \frac{2.40~cm}{\sqrt{2}}$
$z = 1.70~cm$
$E$ is a maximum at a distance of $~~1.70~cm~~$ from the ring's center.
