Figure shows variation of magnetic field inside and outside two cylindrical wires P and Q carrying currents that are uniformly distributed across the cross-section of wires. If current density for two wires are $j_P$ and $j_Q$, then

At the surface of a current-carrying wire,
$$\begin{gathered} B=\frac{1}{2}{\mu }_{0}Jr \\ J\propto \frac{B}{r} \end{gathered}$$
$\therefore$ as slope of wire P > slope of wire Q, therefore $j_P>j_Q$

From the graph, $r_P<r_Q$

$i_Q=j_Q\times \pi r_Q^2$

$i_P=j_P\times \pi r_P^2$

If $j_P{r}_P^2>j_Q{r}_Q^2$, then $i_P>i_Q$

If $j_P{r}_P^2<j_Q{r}_Q^2$, then $i_P<i_Q$

If $j_P{r}_P^2=j_Q{r}_Q^2$, then $i_P=i_Q$