Magnitude of magnetic field (in SI units at the center 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 : 

Magnetic field due to one side at center is 

$\begin{array}{l}B=\frac{{\mu }_{0}i}{4\pi r}\left(\sin \alpha +\sin \beta \right)\text{}=\frac{{\mu }_{0}i}{4\pi \left(\frac{\sqrt{3}a}{2}\right)}\left(\sin 30º+\sin 30º\right) =\frac{{\mu }_{0}i}{2\pi \sqrt{3}a}\text{}\\ There are 6 sides and 50 turns \\ \text{}B=6\times 50\times \frac{{\mu }_{0}i}{2\pi \sqrt{3}\left(0.1\right)}=500\sqrt{3}\frac{{\mu }_{0}i}{\pi }\end{array}$