Suppose third charge is of same nature as that of Q and let it magnitude be q.

So, net force on it is

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=2\mathrm{Fcos\theta } \\ \text{ where }\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}·\frac{\mathrm{Qq}}{\left({\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}\right)}\text{ and }\mathrm{cos\theta }=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}}} \\ \therefore {\mathrm{F}}_{\text{net }}=2\times \frac{1}{4{\mathrm{\pi \epsilon }}_0}·\frac{\mathrm{Qq}}{\left({\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}\right)}\times \frac{\mathrm{x}}{{\left({\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}\right)}^{1/2}} \\ =\frac{2\mathrm{Qqx}}{4{\mathrm{\pi \epsilon }}_0{\left({\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}\right)}^{3/2}} \\ \text{ For }{\mathrm{F}}_{\text{net }}\text{ to be maximum, }\frac{{\mathrm{dF}}_{\text{net }}}{\mathrm{dx}}=0 \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{2\mathrm{Qqx}}{4{\mathrm{\pi \epsilon }}_0{\left({\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}\right)}^{3/2}}\right]=0 \\ \text{ or }\left[{\left({\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}\right)}^{-3/2}-3{\mathrm{x}}^2{\left({\mathrm{x}}^2+\frac{{\mathrm{d}}^2}{4}\right)}^{-5/2}\right]=0 \\ \text{ i.e., }\mathrm{x}=\pm \frac{\mathrm{d}}{2\sqrt{2}} \end{gathered}$$