Magnetic Field of Straight Wires question from JEE Main 2020 (3 Sep, Shift 1)

Answer: (c)

1. Hexagon side $a$: distance of centre from a side $d = \frac{\sqrt{3}a}{2}$, each side subtends $\pm 30^\circ$.
2. One side: $\frac{\mu_0 I}{4\pi d}(2\sin 30^\circ) = \frac{\mu_0 I}{4\pi d}$.
3. $B = 6N\frac{\mu_0 I}{4\pi d} = \frac{300\,\mu_0 I}{4\pi \times 0.05\sqrt{3}} = \frac{1500}{\sqrt{3}}\cdot\frac{\mu_0 I}{\pi}$
4. $B = 500\sqrt{3}\,\frac{\mu_0 I}{\pi}$, so the answer is (c).