Explanation

(b) The system is in equilibrium means the force experienced by each charge is 0. It is clear that charge placed at centre would be in equilibrium for any value of$\ q$, so we are considering the equilibrium of charge placed at any corner.
$F_{CD} + F_{CA}\cos{45{^\circ} + F_{co}\cos{45{^\circ} = 0}}$
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$\frac{1}{4\pi\varepsilon_{0}}.\frac{\left( \mathcal{- Q} \right)\left( \mathcal{- Q} \right)}{a^{2}} + \frac{1}{4\pi\varepsilon_{0}}\frac{\left( \mathcal{- Q} \right)\left( \mathcal{- Q} \right)}{\left( \sqrt{2a} \right)^{2}}$
$\times \frac{1}{\sqrt{2}} + \frac{1}{4\pi\varepsilon_{0}}\frac{\left( \mathcal{- Q} \right)q}{\ \left( \sqrt{2a/2)} \right)^{2}} \times \frac{1}{\sqrt{2}} = 0$
$\frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}^{2}}{a^{2}} + \frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}^{2}}{{2a}^{2}}.\frac{1}{\sqrt{2}} - \frac{1}{4\pi\varepsilon_{0}}\frac{2\mathcal{Q}q}{a^{2}} \times \frac{1}{\sqrt{2}} = 0$
$\mathcal{\Longrightarrow \ \ Q +}\frac{\mathcal{Q}}{2\sqrt{2}} - \sqrt{2q} = 0$
$\Longrightarrow \ 2\sqrt{2}\mathcal{Q + Q -}4q = 0$
$\ \Longrightarrow 4q = (2\sqrt{2} + 1)\mathcal{Q}$
$\Longrightarrow q = \left( 2\sqrt{2} + 1 \right)\frac{\mathcal{Q}}{4}\ $