The total vector of magnetic induction $\vec{B}$ is found as the sum of the vectors of magnetic induction created by each of the wires $$\vec{B} = \vec{B_1} + \vec{B_2} + \vec{B_3} +\vec{B_4}$$ Since $\vec{B_2}$ and $\vec{B_4}$ are oppositely directed - equal in magnitude and opposite in direction, then the total magnetic induction vector $\vec{B}$ will be co-directed with $\vec{B_1}$ and $\vec{B_3}$ and equal to the sum of their moduli $$\vec{B_2} = -\vec{B_4} \Rightarrow \boxed{B = B_1 + B_3 = 2B_1}\tag{1}$$ We find the magnetic induction created by one wire at a distance of $a/\sqrt{2}$ from the center of the square as $$B_1 = \frac{\mu_0 I}{2\pi\frac{a}{\sqrt{2}}} = \frac{\mu_0 I \sqrt{2}}{2\pi a}$$ Next, we substitute the obtained expression for $B_1$ into the expression $(1)$ for the total induction of the described system $$\boxed{B = \frac{\mu_0 I \sqrt{2}}{\pi a} = 18.9~\mu T}$$