Answer
$E = 0$
Work Step by Step
We can find the magnitude of the electric field at $z = \infty$:
$E = \frac{Qz}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}}$
$E \approx \frac{Qz}{4\pi~\epsilon_0~(z^2)^{3/2}}$
$E \approx \frac{Qz}{4\pi~\epsilon_0~z^3}$
$E \approx \frac{Q}{4\pi~\epsilon_0~z^2}$
$E = 0~~$ since $~~z =\infty$
