Answer
$B=4.00 \times 10^{-3} \mathrm{~T} $
Work Step by Step
We use Ampere's law: $\oint \vec{B} \cdot d \vec{s}=\mu_0 i$, where the integral is around a closed loop and $i$ is the net current of the loop. The magnitude of magnetic field is
$$
B=\frac{\mu_0 i}{2 \pi r}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(25.0 \mathrm{~A})}{2 \pi\left(1.25 \times 10^{-3} \mathrm{~m}\right)}\\=4.00 \times 10^{-3} \mathrm{~T} .
$$
