JEE Main · 2007 · 29 Apr 2007 · Offline · Magnetic Effects of Current

Treat the current as uniformly spread over the cross-section and apply Ampere's law on circular loops. Inside the wire at $r=\frac{a}{2}$ only the enclosed fraction of the current contributes, giving $B_{\mathrm{in}}=\frac{{\mu }_{0}i}{2\pi a^2}\cdot \frac{a}{2}=\frac{{\mu }_{0}i}{4\pi a}$. Outside at $r=2a$ the whole current is enclosed, giving $B_{\mathrm{out}}=\frac{{\mu }_{0}i}{2\pi (2a)}=\frac{{\mu }_{0}i}{4\pi a}$. The two fields come out equal, so their ratio is $1$.