where, \(a\) is the radius of loop.

Then, \(B_{1}=\frac{\mu_{0} I}{2 a}\)

Now, for coil \(\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi} \cdot \frac{2 \mathrm{nA}}{\mathrm{x}^{3}}\)

at the centre \(x=\) radius of loop

\(\mathrm{B}_{2} =\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \times 3 \times(1 / 3) \times \pi(\mathrm{a} / 3)^{2}}{(\mathrm{a} / 3)^{3}} \)

\(=\frac{\mu_{0} \cdot 3 \mathrm{I}}{2 \mathrm{a}} \)

\(\therefore \frac{B_{1}}{B_{2}}=\frac{\mu_{0} I / 2 a}{\mu_{0} \cdot 3I / 2 a}\)

\(B_{1}: B_{2}=1: 3\)