From ampere’s circuital law, for a long straight wire of circular cross section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is 

Magnetic field due to a thick circular cross section rod can be calculated by Ampere’s circuital law.
If total current is I, then current per unit area is $\frac{\mathrm{I}}{\pi {\mathrm{R}}^2}$ where R is the radius of the rod.
Considering a point inside, at a distance ‘r’ from the centre, current enclosed in the region, $\mathrm{i}=\frac{\mathrm{I}}{\pi {\mathrm{R}}^2}\pi {\mathrm{r}}^2$.
According to Ampere’s circuital law,
$\begin{array}{l}\text{∮}B\cdot d\ell ={\mu }_{0}i\\ B2\pi r={\mu }_0\frac{I}{\pi R^2}\pi r^2\\ B=\frac{{\mu }_{0}I\cdot r}{2\pi R^2}\end{array}$
So, B is directly proportional to r inside the conductor.
If the point is outside, $\text{∮}\mathrm{B}\cdot \mathrm{d}\ell ={\mu }_0\mathrm{I},$as total current is enclosed.
$B=\frac{{\mu }_{0}l}{2\pi r}$
Hence, B is inversely proportional to distance r.