The magnetic field due to the current-carrying finite length wire is given by

$B=\frac{{\mu }_{0}I}{4\pi R}\left[\sin {\theta }_1+\sin {\theta }_2\right]$

Here for hexagonal shape coil, ${\theta }_1={\theta }_2=30°$.

So, the net magnetic field at the centre will be $\mathrm{B}=50\times 6\times \frac{{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi }\left({\displaystyle \frac{10}{100}}\cos 30°\right)}\left[\sin 30°+\sin 30°\right]$

$B=2\times 75\times 10\frac{{\mathrm{\mu }}_0\mathrm{i}}{\sqrt{3}\mathrm{\pi }}\left(\frac{1}{2}+\frac{1}{2}\right)$

$$\begin{gathered} B=\frac{1500}{\sqrt{3}}\frac{{\mu }_{0}i}{\mathrm{\pi }} \\ =\left(500\sqrt{3}\right)\frac{{\mu }_{0}i}{\mathrm{\pi }} \end{gathered}$$