An electric current flows along the length of a hollow tube,...

By Ampere's circuital law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$.

For any point inside the hollow tube at a distance $r$ from the axis ($r < R$), the Amperian loop encloses no current, so $I_{enclosed} = 0$.

$B(2\pi r) = 0 \Rightarrow B = 0$

Thus, the magnetic field is zero at all points inside the tube, including the axis. Since it is zero everywhere inside, it remains constant.

For a point outside the tube ($r > R$), the enclosed current is the total current $I$.

$B(2\pi r) = \mu_0 I \Rightarrow B = \dfrac{\mu_0 I}{2\pi r}$

Thus, the magnetic field is non-zero immediately outside the tube and decreases as $\dfrac{1}{r}$.