Answer
$\theta = 54.7^{\circ}$
Work Step by Step
For each $l$, the values of $m_l$ can be $~~m_l = 0, \pm 1, \pm 2,...,\pm l$
The second largest possible value of $m_l$ is $~~m_l = 2$
$L= \sqrt{12}~\hbar$
$L_z = m_l~\hbar = 2~\hbar$
We can find the value for $\theta$:
$cos~\theta = \frac{L_z}{L}$
$cos~\theta = \frac{2~\hbar}{\sqrt{12}~\hbar}$
$cos~\theta = \frac{2}{\sqrt{12}}$
$\theta = cos^{-1}~(\frac{2}{\sqrt{12}})$
$\theta = 54.7^{\circ}$
