Answer
$E=\frac{3 Q}{4 \pi \varepsilon_0 z^4}$.
Work Step by Step
Use the binomial expansions
$
\begin{aligned}
& (z-d / 2)^{-3} \approx z^{-3}-3 z^{-4}(-d / 2) \\
& (z+d / 2)^{-3} \approx z^{-3}-3 z^{-4}(d / 2)
\end{aligned}
$$
we obtain
$$
E=\frac{q d}{2 \pi \varepsilon_0(z-d / 2)^3}-\frac{q d}{2 \pi \varepsilon_0(z+d / 2)^3} \\\approx \frac{q d}{2 \pi \varepsilon_0}\left[\frac{1}{z^3}+\frac{3 d}{2 z^4}-\frac{1}{z^3}+\frac{3 d}{2 z^4}\right]\\=\frac{6 q d^2}{4 \pi \varepsilon_0 z^4} .
$
Since the quadrupole moment is $Q=2 q d^2$, we have $E=\frac{3 Q}{4 \pi \varepsilon_0 z^4}$.
