Answer
$B=6.11 \times 10^{-5} \mathrm{~T} .$
Work Step by Step
The angle between the magnetic axis and the rotational axis of the Earth is $11.5^{\circ}$, so $\lambda_m=90.0^{\circ}-11.5^{\circ}=78.5^{\circ}$ at Earth's geographic north pole. Also $r=R_e=6370 \mathrm{~km}$. Thus,
$
\begin{aligned}
B & =\frac{\mu_0 \mu}{4 \pi R_E^3} \sqrt{1+3 \sin ^2 \lambda_m}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)\left(8.0 \times 10^{22} \mathrm{~J} / \mathrm{T}\right) \sqrt{1+3 \sin ^2 78.5^{\circ}}}{4 \pi\left(6.37 \times 10^6 \mathrm{~m}\right)^3} \\
& =6.11 \times 10^{-5} \mathrm{~T} .
\end{aligned}
$
