Answer
$B_{}=6.3×10−7 T$
Work Step by Step
We use $\oint \vec{B} \cdot d \vec{s}=\mu_0 I_{\text {enclosed }}$ to find
$
\begin{aligned}
B & =\frac{\mu_0 I_{\text {enchosed }}}{2 \pi r}=\frac{\mu_0\left(J_d \pi r^2\right)}{2 \pi r}=\frac{1}{2} \mu_0 J_d r=\frac{1}{2}\left(1.26 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)\left(20 \mathrm{~A} / \mathrm{m}^2\right)\left(50 \times 10^{-3} \mathrm{~m}\right) \\
& =6.3 \times 10^{-7} \mathrm{~T}
\end{aligned}
$
