For a finite wire of length $2a$ at perpendicular distance $a$ from centre:
$$B_{side} = \frac{\mu_0 I}{4\pi a}(\sin 45° + \sin 45°) = \frac{\mu_0 I}{4\pi a} \cdot \sqrt{2}$$
$$B_{side} = \frac{\mu_0 I\sqrt{2}}{4\pi a}$$

All 4 sides contribute equally and fields add up (by right-hand rule, all point in the same direction):
$$B_{total} = 4 \times \frac{\mu_0 I\sqrt{2}}{4\pi a} = \frac{\sqrt{2}\,\mu_0 I}{\pi a}$$