 Solution

Let the wire radius be $a$, and take its electric current along $+z$. Cylindrical symmetry makes the magnetic flux density azimuthal. For a circular Amperian loop of radius $r$, Ampère's law gives $2\pi rB={\mu }_0{I}_{\mathrm{enclosed}}$. Uniform current density means $I_{\mathrm{enclosed}}=I{r}^2/a^2$ inside and $I$ outside. ConsequentlyIts direction follows the right-hand rule. The magnitude rises linearly from zero to ${\mu }_{0}I/(2\pi a)$ at the surface, is continuous there, and then falls as $1/r$.