For the entire system to be in equilibrium, the net force on every charge must be zero. By symmetry, the force on the central charge $q$ is already zero. We must ensure the net force on any corner charge $Q$ is zero.

Consider the charge $Q$ at one corner. The forces acting on it are:

1. Repulsion from two adjacent corner charges (distance $a$): Resultant $F_1 = \sqrt{2} \frac{kQ^2}{a^2}$ directed along the diagonal outward.
2. Repulsion from the opposite corner charge (distance $\sqrt{2}a$): $F_2 = \frac{kQ^2}{(\sqrt{2}a)^2} = \frac{kQ^2}{2a^2}$ directed along the diagonal outward.
3. Force from the center charge $q$ (distance $a/\sqrt{2}$): $F_q = \frac{kQq}{(a/\sqrt{2})^2} = \frac{2kQq}{a^2}$ directed along the diagonal.

For equilibrium ($F_{net} = 0$): $F_1 + F_2 + F_q = 0$ $\frac{kQ^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) + \frac{2kQq}{a^2} = 0$ $Q \left( \frac{2\sqrt{2} + 1}{2} \right) + 2q = 0$ $2q = - \frac{Q}{2} (1 + 2\sqrt{2})$ $q = - \frac{Q}{4} (1 + 2\sqrt{2})$

(See NCERT Physics Class 12, Chapter 1, for principles of superposition and Coulomb's Law).