By Ampere's circuital law, for a closed loop inside the hollow tube:
$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$$
Since the tube is thin-walled, no current is enclosed inside the tube.
$$I_{enclosed} = 0 \Rightarrow B = 0$$
The magnetic field inside a long straight thin-walled hollow current-carrying tube is zero.