Answer
$B=6.20 \times 10^{-5} \mathrm{~T}$
Work Step by Step
At the north magnetic pole $\left(\lambda_m=90.0^{\circ}\right)$, we obtain
$$
B=\frac{\mu_0 \mu}{4 \pi r^3} \sqrt{1+3 \sin ^2 \lambda_m}\\=\left(3.10 \times 10^{-5}\right) \sqrt{1+3(1.00)^2}\\=6.20 \times 10^{-5} \mathrm{~T} .
$$
