Consider the forces acting on one of the vertex charges, say at the top vertex. The other two vertex charges (both $-q$) exert repulsive forces on it. Let the magnitude of each of these forces be $F_q$.

$$F_q = k \frac{(-q)(-q)}{r^2} = k\frac{q^2}{r^2}$$

The angle between these two force vectors is $60^\circ$. The magnitude of their resultant force $F_{res}$ is:

$$F_{res} = \sqrt{F_q^2 + F_q^2 + 2 F_q F_q \cos(60^\circ)} = \sqrt{3} F_q = \sqrt{3} k \frac{q^2}{r^2}$$

This resultant force points directly away from the centroid. For equilibrium, the force exerted by the central charge $Q$, denoted as $F_Q$, must be equal in magnitude and opposite in direction. The distance $d$ from the centroid to any vertex of an equilateral triangle with side $r$ is $d = r/\sqrt{3}$.

$$F_Q = k \frac{Q(-q)}{d^2} = k \frac{-qQ}{(r/\sqrt{3})^2} = -3k \frac{qQ}{r^2}$$

The net force is zero, so $F_{res} + F_Q = 0$.

$$\sqrt{3} k \frac{q^2}{r^2} - 3k \frac{qQ}{r^2} = 0$$ $$\sqrt{3} q = 3Q$$ $$Q = \frac{\sqrt{3}}{3}q = \frac{q}{\sqrt{3}}$$