Answer
$B = 8.50\times 10^{-4}~T$
Work Step by Step
We can find the enclosed current:
$i_{enc} = \frac{(\pi)~(1.00~cm)^2}{(\pi)(2.00~cm)^2}~(170~A) = 42.5~A$
We can find the magnitude of the magnetic field:
$\int~B\cdot ds = \mu_0~i_{enc}$
$B\cdot 2\pi~r = \mu_0~i_{enc}$
$B = \frac{\mu_0~i_{enc}}{2\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)(42.5~A)}{(2\pi)(0.0100~m)}$
$B = 8.50\times 10^{-4}~T$
