Answer
The magnitude of the current density in wire $a$ is greater than that in wire $c$
Work Step by Step
$\int~B\cdot ds = \mu_0~i_{enc}$
Therefore:
$B = \frac{\mu_0~i_{enc}}{2\pi~r}~~~$ where $r$ is the radial distance from the center of the wire.
At large values of $r$ on the graph, $a$ and $c$ have the same magnitude of magnetic fields.
Therefore, the enclosed current in curves $a$ and $c$ must be equal. This shows that wire $a$ and wire $c$ have the same current.
Each curve changes from increasing to decreasing at the point where $r$ is the radius of each wire.
On the graph, we can see that wire $a$ has a smaller radius than wire $c$
Therefore, wire $a$ has a greater current density than wire $c$
The magnitude of the current density in wire $a$ is greater than that in wire $c$
