Answer
$J_a$ maximizes the current density near the wire’s surface.
Work Step by Step
$J_a = \frac{J_0~r}{R}$
As $r$ approaches $R$, then $J_a$ approaches $J_0 = 5.50\times 10^4~A/m^2$
$J_b = J_0~(1-\frac{r}{R})$
As $r$ approaches $R$, then $J_b$ approaches $0$
Therefore, $J_a$ maximizes the current density near the wire’s surface.
