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

Using Ampere's circuital law:

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

For a point inside the hollow cylindrical tube at a distance $r/2$ from the surface, the distance from the axis is $r - r/2 = r/2$.

An Amperian loop of radius $r/2$ inside the hollow tube encloses no current, so $I_{enclosed} = 0$.

Therefore, $B \times 2\pi \left(\dfrac{r}{2}\right) = 0 \Rightarrow B = 0$

The magnetic field at any point inside the hollow tube is zero.