Answer
$B=5.59 \times 10^{-5} \mathrm{~T} $
Work Step by Step
At $\lambda_m=60.0^{\circ}$, we find
$$
B=\frac{\mu_0 \mu}{4 \pi r^3} \sqrt{1+3 \sin ^2 \lambda_m}\\=\left(3.10 \times 10^{-5}\right) \sqrt{1+3 \sin ^2 60.0^{\circ}}\\=5.59 \times 10^{-5} \mathrm{~T} .
$$
