A long, thin, hollow cylindrical tube of radius r carries a current...

Let the distance of the point from the axis of the cylinder be $R$.

Since the point is at a distance $r/2$ from the surface outside the tube, we have $R = r + \dfrac{r}{2} = \dfrac{3r}{2}$.

By Ampere's circuital law, the magnetic field outside a long cylindrical tube carrying current $i$ is given by:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 i$

$B(2\pi R) = \mu_0 i$

$B = \dfrac{\mu_0 i}{2\pi R}$

Substituting $R = \dfrac{3r}{2}$:

$B = \dfrac{\mu_0 i}{2\pi \left(\dfrac{3r}{2}\right)} = \dfrac{\mu_0 i}{3\pi r}$