Given here, $N=100$, $R=5 \mathrm{cm}$ and $B=37.68\times {10}^{-4} \mathrm{T}$

Magnetic field at the centre of circular coil is given by $B_{\mathrm{centre} }=\frac{N{\mu }_{0}l}{2R}$.

Putting the values, 

$37.68\times {10}^{-4}=\frac{100\times 4\pi \times {10}^{-7}\times I}{2\times 5\times {10}^{-2}}$

Current through coil is $I=3 \mathrm{A}$.