Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm, 50 turns and carrying current $$I$$ (Ampere) in units of $$\frac{\mu_0 I}{\pi}$$ is:

Perpendicular distance from the centre to each side of a regular hexagon:

$$d = a \cos(30^\circ) = 0.1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{20}\ \text{m}$$

Angles subtended by each side at the centre: $$\theta_1 = \theta_2 = 30^\circ$$

Magnetic field due to one side of a single turn:

$$B_1 = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2) = \frac{\mu_0 I}{4\pi \left(\frac{\sqrt{3}}{20}\right)}(\sin 30^\circ + \sin 30^\circ)$$

$$B_1 = \frac{20\mu_0 I}{4\pi \sqrt{3}}\left(\frac{1}{2} + \frac{1}{2}\right) = \frac{5\mu_0 I}{\pi \sqrt{3}}$$

$$B = 6 \times n \times B_1 = 6 \times 50 \times \frac{5\mu_0 I}{\pi \sqrt{3}}$$

$$B = \frac{1500}{\sqrt{3}}\frac{\mu_0 I}{\pi} = 500\sqrt{3}\left(\frac{\mu_0 I}{\pi}\right)$$