= Solution
Set the normal <electric field> to zero and cancel its nonzero overall factor:
$$
\frac4{(x^2+4d^2)^{3/2}}=\frac1{(x^2+d^2)^{3/2}}.
$$
Taking the power $2/3$ gives $4^{2/3}(x^2+d^2)=x^2+4d^2$, hence
$$
\boxed{x=d\sqrt{\frac{4-4^{2/3}}{4^{2/3}-1}}\simeq0.98686d=0.98686\ \mathrm m}.
$$
The numerical value is independent of the nonzero magnitude of $Q$. Beyond this radius the signed normal <electric field> is positive. Thus its magnitude first falls to zero, rises again and eventually decays; it is not globally monotone.
For $x\gg d$, expanding each denominator gives
$$
E_z=\frac{Qd}{2\pi\epsilon_0x^3}\left[3-\frac{45}{2}\frac{d^2}{x^2}+O\!\left(\frac{d^4}{x^4}\right)\right],\qquad\boxed{E_z\sim\frac{3Qd}{2\pi\epsilon_0x^3}>0}.
$$
This is a dipole rather than a monopole tail. The real-plus-image charge distribution has zero total charge and dipole moment $p_z=-6Qd$; in its equatorial plane a dipole produces $E_z=-k_ep_z/x^3$, confirming the coefficient and sign.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016-image-charge-surface-field.png]
{title=Image charges and the sign-changing normal field above an infinite conducting plane}
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