Answer
$B=\frac{\mu_0 \mu}{4 \pi r^3} \sqrt{1+3 \sin ^2 \lambda_m} \text {}$
Work Step by Step
Substituting the expression given in the problem statement, we have
$
\begin{aligned}
B & =\sqrt{B_h^2+B_v^2}=\sqrt{\left(\frac{\mu_0 \mu}{4 \pi r^3} \cos \lambda_m\right)^2+\left(\frac{\mu_0 \mu}{2 \pi r^3} \sin \lambda_m\right)^2}=\frac{\mu_0 \mu}{4 \pi r^3} \sqrt{\cos ^2 \lambda_m+4 \sin ^2 \lambda_m} \\
& =\frac{\mu_0 \mu}{4 \pi r^3} \sqrt{1+3 \sin ^2 \lambda_m},
\end{aligned}
$
where $\cos ^2 \lambda_m+\sin ^2 \lambda_m=1$ was used.
