Answer
$z = 0.346~m$
Work Step by Step
We can write an expression for the electric field due to a disk of charge:
$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})$
We can find an expression for the electric field at the center of the disk:
$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})$
$E = \frac{\sigma}{2\epsilon_0}(1-\frac{0}{\sqrt{0^2+R^2}})$
$E = \frac{\sigma}{2\epsilon_0}$
We can find $z$ when $~~E = \frac{1}{2}\times \frac{\sigma}{2\epsilon_0}$:
$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}}) = \frac{1}{2}\times \frac{\sigma}{2\epsilon_0}$
$(1-\frac{z}{\sqrt{z^2+R^2}}) = \frac{1}{2}$
$\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{2}$
$2z = \sqrt{z^2+R^2}$
$4z^2 = z^2+R^2$
$3z^2 = R^2$
$z = \frac{R}{\sqrt{3}}$
$z = \frac{0.600~m}{\sqrt{3}}$
$z = 0.346~m$
