Answer
$\mu=3.46\mu_B$
Work Step by Step
The magnitude of the orbital magnetic moment of the electron given by
$\mu=\frac{e\hbar}{2m}\sqrt {l(l+1)}=\mu_B\sqrt {l(l+1)}$
where, $\mu_B=\frac{e\hbar}{2m}$ is the Bohr magneton.
For, $l=3$
$\mu=\mu_B\sqrt {3(3+1)}=3.46\mu_B$
