Using ampere circuital law, $\oint \overrightarrow{B}·\overrightarrow{dl}={\mu }_{0}I$

If an ampere loop is taken inside the hollow cylinder then, $\oint \overrightarrow{B}·\overrightarrow{dl}={\mu }_0\times 0=0$

If an ampere loop is taken of radius $r$such that $r\ge R$then, 

$$\begin{gathered} \oint \overrightarrow{B}·\overrightarrow{dl}={\mu }_{0}I \\ B\left(2\mathrm{\pi r}\right)={\mu }_{0}I \\ B=\frac{{\mu }_{0}I}{2\mathrm{\pi r}} \end{gathered}$$

As it can be clearly seen that there is no magnetic field inside the hollow cylindrical metal and outside the cylinder $B\propto \frac{1}{r}$