Answer
$B = 4.0\times 10^{-7}~T$
Work Step by Step
We can find the enclosed current:
$i_{enc} = \int_{0}^{a}~\frac{2\pi~r~J_0~r~dr}{a}$
$i_{enc} = \int_{0}^{a}~\frac{2\pi~r^2~J_0~dr}{a}$
$i_{enc} = \frac{2~J_0~\pi~r^3}{3a} \Big \vert_{0}^{a}$
$i_{enc} = \frac{2~J_0~\pi~a^3}{3a}-0$
$i_{enc} = \frac{2~J_0~\pi~a^2}{3}$
$i_{enc} = \frac{(2)(310~A/m^2)~(\pi)~(0.0031~m)^2}{3}$
$i_{enc} = 6.24~mA$
We can find the magnitude of the magnetic field at $r = a$:
$\int~B\cdot ds = \mu_0~i_{enc}$
$B\cdot 2\pi~r = \mu_0~i_{enc}$
$B = \frac{\mu_0~i_{enc}}{2\pi~a}$
$B = \frac{(4\pi\times 10^{-7}~H/m)(6.24~mA)}{(2\pi)(0.0031~m)}$
$B = 4.0\times 10^{-7}~T$
