The correct option of this question will be (c).

**Solution —**

Current density, $J=\frac{I}{\pi a^{2}}$

From Ampere's circuital law

$\oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} \cdot I_{\text {enclosed }}$

For $r<a, B \times 2 \pi r=\mu_{0} \times J \times \pi r^{2}$

$\Rightarrow B=\frac{\mu_{0} I}{\pi a^{2}} \times \frac{r}{2}$

At $r=a / 2, B_{1}=\frac{\mu_{0} I}{4 \pi a}$

For $r>a, B \times 2 \pi r=\mu_{0} I \Rightarrow B=\frac{\mu_{0} I}{2 \pi r}$

At $r=2 a, B_{2}=\frac{\mu_{0} I}{4 \pi a}$

So, $\frac{B_{1}}{B_{2}}=1$