Answer
$B=3.10 \times 10^{-5} \mathrm{~T} \text {}$
Work Step by Step
$
\text { The inclination } \phi_i \text { is related to } \lambda_m \text { by } \tan \phi_i=\frac{B_v}{B_h}\\=\frac{\left(\mu_0 \mu / 2 \pi r^3\right) \sin \lambda_m}{\left(\mu_0 \mu / 4 \pi r^3\right) \cos \lambda_m}\\=2 \tan \lambda_m \text {. }
$
At the magnetic equator $\left(\lambda_m=0\right), \phi_i=0^{\circ}$, and the field is
$
B=\frac{\mu_0 \mu}{4 \pi r^3}\\=\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(6.37 \times 10^6 \mathrm{~m}\right)^3}\\=3.10 \times 10^{-5} \mathrm{~T} \text {. }
$
