The charge $\mathrm{A}$ experiences three forces ${\mathrm{F}}_1$, ${\mathrm{F}}_2$ and ${\mathrm{F}}_3$ as shown in the figure.



Now, ${\mathrm{F}}_1=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{ }}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{a}}^2}$ (along $\mathrm{AB}$)

${\mathrm{F}}_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{ }}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{a}}^2}$ (along $\mathrm{AD}$)

${\mathrm{F}}_3=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{ }}\frac{{\mathrm{q}}_1^2}{{\mathrm{a}}^2}$

The resultant of ${\mathrm{F}}_1$ and ${\mathrm{F}}_2$ should be equal and opposite to ${\mathrm{F}}_3$ to keep the system in equilibrium.

Resultant of $F_1$ and ${\mathrm{F}}_2$$={\mathrm{F}}_{\mathrm{R}}=\sqrt{{\mathrm{F}}_1^2+\mathrm{ }{\mathrm{F}}_2^2}$

$=\frac{1}{4\mathrm{\pi }\mathrm{ }}\frac{{\mathrm{q}}_1{\mathrm{q}}_1}{{\mathrm{\epsilon }}_0{\mathrm{a}}^2}\sqrt{2}$

For equilibrium,

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{a}}^2}\sqrt{2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1^2}{{\mathrm{a}}^2}$

$\frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=\sqrt{2}$