Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be

Maximum possible magnetic field is at the surface
\(\begin{aligned}
& \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\
& \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}
\end{aligned}\)
It can be obtained inside as well as outside the wire For inside,
\(\begin{aligned}
& \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{Ir}}{2 \pi \mathrm{a}^2} \\
& \Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2}
\end{aligned}\)
For outside
\(\begin{aligned}
& \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\
& \Rightarrow \mathrm{r}=2 \mathrm{a}
\end{aligned}\)

Correct answer \(\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]\)