Answer
$\theta = 80.4^{\circ}$
Work Step by Step
We can find the required speed at the top of the loop in order for the string not to go slack:
$\frac{mv^2}{r} = mg$
$v^2 = gr$
$v = \sqrt{gr}$
$v = \sqrt{\frac{g~L}{3}}$
We can use conservation of energy to find the initial height:
$U_1 = U_2+K_2$
$mgh_1 = mgh_2+\frac{1}{2}mv^2$
$gh_1 = gh_2+\frac{1}{2}(\sqrt{\frac{g~L}{3}})^2$
$gh_1 = \frac{2gL}{3}+\frac{1}{2}(\frac{g~L}{3})$
$h_1 = \frac{2L}{3}+\frac{L}{6}$
$h_1 = \frac{5L}{6}$
We can find the angle $\theta$:
$cos~\theta = \frac{L/6}{L}$
$cos~\theta = \frac{1}{6}$
$\theta = cos^{-1}~(\frac{1}{6})$
$\theta = 80.4^{\circ}$
