A long thin hollow metallic cylinder of radius R has a current i...

Using ampere circuital law, $∮ \overset{\to}{B} \cdot \overset{\to}{d l} = {\mu}_{0}I$

If an ampere loop is taken inside the hollow cylinder then, $∮ \overset{\to}{B} \cdot \overset{\to}{d l} = {\mu}_{0} \times 0 = 0$

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

$∮ \overset{\to}{B} \cdot \overset{\to}{d l} = {\mu}_{0}I$
$B\left(2\pi r\right) = {\mu}_{0}I$
$B = \dfrac{{\mu}_{0}I}{2\pi r}$

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