b
For loop $\mathrm{B}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{a}}$
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}} $
$\therefore \frac{B_{1}}{B_{2}}=\frac{\mu_{0} I / 2 a}{\mu_{0} \cdot 3I / 2 a}$
$B_{1}: B_{2}=1: 3$