Explanation

(c) Current density $J = \frac{i}{\pi a^{2}}$
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From Ampere’s circuital law
$\oint_{}^{}{B.dl = \mu_{0}\ .i_{enclosed}}$
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 $r = \frac{a}{2}$
$B_{1} = \frac{\mu_{0}i}{4\pi a}$
For $r > a$
$B \times 2\pi r = \mu_{0}i\ \Rightarrow B = \ \frac{\mu_{0}i}{2\pi r}$
At $r = 2a,\ \ \ \ \ \ \ \ B_{2} = \frac{\mu_{0}i}{4\pi a}$
So, $\frac{B_{1}}{B_{2}} = 1$