Answer
$B_{max}=3.83 \times 10^{-4} \mathrm{~T} .$
Work Step by Step
For maximum $B$, we set $\sin \lambda_{\mathrm{m}}=1.00$. Also, $r=6370 \mathrm{~km}-2900 \mathrm{~km}=3470 \mathrm{~km}$. Thus,
$
\begin{aligned}
B_{\max } & =\frac{\mu_0 \mu}{4 \pi r^3} \sqrt{1+3 \sin ^2 \lambda_m}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)\left(8.00 \times 10^{22} \mathrm{~A} \cdot \mathrm{m}^2\right)}{4 \pi\left(3.47 \times 10^6 \mathrm{~m}\right)^3} \sqrt{1+3(1.00)^2} \\
& =3.83 \times 10^{-4} \mathrm{~T} .
\end{aligned}
$
