A long straight wire of radius 'r' carries a steady current 'I'. The current is uniformly distributed over its cross-section. The ratio $\left(\frac{B}{B^1}\right)$ of the magnetic field ' $B$ ' and ' $B^1$ ' it radial distances $\frac{r}{2}$ ' and ' $3 r$ ' respectively, from the axis of the wire is

For a long straight wire carrying uniform current $I$ with radius $r$, the magnetic field varies with radial distance $r'$ from the axis. The field inside the wire ($r' < r$) follows $B = \frac{\mu_0 I r'}{2\pi r^2}$, while outside ($r' > r$) it follows $B = \frac{\mu_0 I}{2\pi r'}$.

At distance $r' = r/2$ (inside the wire):

$B = \frac{\mu_0 I (r/2)}{2\pi r^2} = \frac{\mu_0 I}{4\pi r}$

At distance $r' = 3r$ (outside the wire):

$B' = \frac{\mu_0 I}{2\pi (3r)} = \frac{\mu_0 I}{6\pi r}$

The ratio is then:

$\frac{B}{B'} = \frac{\frac{\mu_0 I}{4\pi r}}{\frac{\mu_0 I}{6\pi r}} = \frac{1/4}{1/6} = \frac{3}{2}$

Final answer: $\boxed{B}$