Answer
$B = 7.7~mT$
Work Step by Step
We can find the magnetic field at the surface of the wire:
$\int~B\cdot ds = \mu_0~i_{enc}$
$B\cdot 2\pi~r = \mu_0~i_{enc}$
$B = \frac{\mu_0~i_{enc}}{2\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)(50~A)}{(2\pi)(1.3\times 10^{-3}~m)}$
$B = 7.7~mT$
