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