Hint: Use Ampere’s law
Sol: As $B\left(r\right)=B_0{r}^2$



$\therefore$ Total current (upto R)
$B_0{R}^{2}2\pi R={\mu }_0{I}_0 \Rightarrow I_0=\frac{2\pi R^3{B}_0}{{\mu }_0}$
And let upto 'r' it confines half of the current.

Then $B_0{r\text{'}}^2·2\pi r\text{'}={\mu }_0\frac{I_0}{2}$
$\Rightarrow B_0·2\pi {r\text{'}}^3=\frac{{\mu }_0}{2}·\frac{2\pi R^3{B}_0}{{\mu }_0}$
$\Rightarrow r\text{'}=\frac{R}{{\left(2\right)}^{\frac{1}{3}}}$

Now let $\overrightarrow{J}\mathrm{(}r\mathrm{)}$ is the current density at ‘r’
Then $B_0{r}^2·2\pi r={\mu }_0{\int }_0^{r}J\mathrm{(}r\mathrm{)} 2\pi rdr$
$\because \left({\int }_{}^{}Bdl = {\mu }_{0}I\right)$
$\Rightarrow 2\pi B_0{r}^3={\mu }_0·2\pi {\int }_0^{r}J\mathrm{(}r\mathrm{)}rdr$

$\therefore$ Differentiating both side w.r.t(r)
$2\pi B_0·3r^2·dr={\mu }_{0}2\pi ·J\mathrm{(}r\mathrm{)} rdr$
$\Rightarrow J\mathrm{(}r\mathrm{)}=\frac{3B_{0}r}{{\mu }_0}$

For $r>R$;

$B\times 2\pi r={\mu }_0{I}_0$

$\Rightarrow B\times 2\pi r={\mu }_0\times \frac{2\pi R^3{B}_0}{{\mu }_0}$

$\Rightarrow B=\frac{B_0{R}^3}{r}$