A long straight wire of radius ‘a’ carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic fields at distances a/2 and 2a from the axis of wire is

The magnetic field at a point inside the wire and at a distance $r\left(=\frac{{\displaystyle a}}{{\displaystyle 2}}\right)$ from the axis of wire is $B=\frac{{\mu }_{0}I}{2\pi a^2}r=\frac{{\mu }_{0}I}{2\pi a^2}\times \frac{a}{2}=\frac{{\mu }_{0}I}{4\pi a}$
The magnetic field at a point outside the wire at distance r (= 2a) from the axis of wire is 
$B^{\text{'}}=\frac{{\mu }_{0}I}{2\pi r}=\frac{{\mu }_{0}I}{2\pi }\times \frac{1}{2a}=\frac{{\mu }_{0}I}{4\pi a}\therefore \frac{B}{B^{\text{'}}}=1$