Moving Charges and Magnetism question for NEET / JEE Main, with solution

Answer: (D) $\dfrac{2\sqrt{2}\,\mu_0 I}{\pi a}$

Each side is a finite wire at perpendicular distance $\dfrac{a}{2}$ from the centre, and its ends subtend $45^\circ$ on either side:
$$B_1 = \frac{\mu_0 I}{4\pi (a/2)}(\sin 45^\circ + \sin 45^\circ) = \frac{\sqrt{2}\,\mu_0 I}{2\pi a}$$

All four sides give fields in the same direction:
$$B = 4B_1 = \frac{2\sqrt{2}\,\mu_0 I}{\pi a}$$

Solution by Sreeraj P, M.Sc Physics