The distance from each corner to the center is $r = \frac{a\sqrt{2}}{2}$, so $r^2 = a^2/2$. Let the central charge be $q_0$. By superposition, we sum the forces vectorially.

Forces from the diagonal pair $q_2=2q$ and $q_4=2q$ are equal in magnitude and opposite in direction. They cancel each other out.

$$F_2 = k \frac{2q q_0}{r^2}, \quad F_4 = k \frac{2q q_0}{r^2} \implies \vec{F}_2 + \vec{F}_4 = 0$$

Forces from the diagonal pair $q_1=q$ and $q_3=-4q$ are along the same line. The force from $q_1$ is repulsive, and the force from $q_3$ is attractive. Both forces point in the same direction, from $q_1$ towards $q_3$.

$$F_{net} = F_1 + F_3 = k \frac{|q_1| q_0}{r^2} + k \frac{|q_3| q_0}{r^2}$$ $$F_{net} = \frac{k q_0}{r^2} (|q| + |-4q|) = \frac{k q_0}{a^2/2} (5q) = \frac{10k q q_0}{a^2}$$

Assuming a unit positive charge means $q_0 = +1$ C.

$$F_{net} = \frac{10kq}{a^2}$$

The direction is along the diagonal from charge $q$ towards charge $-4q$.