Answer
$80.4^\circ$
Work Step by Step
We need to find the velocity of the ball when it is $\frac{2}{3}L$ from the bottom that will make it complete the vertical circle without falling down.
Then, we can find the angle of release $\theta$ since this potential energy will be transferred to kinetic energy.
We chose our origin point to be the ball's second position, as you see in the figure below.
The system here is the ball+ the Earth.
We need to find the minimum velocity $v_3$ of the ball that allows it to complete the vertical circle (the second circle).
At the top of its small circle, the forces exerted on it are the tension force in the wire and the gravitational force. Thus, applying Newton's second law,
$$\sum F_y=mg+T=ma_r=m\dfrac{v_3^2}{R}$$
where $R=L/3$
$$mg+T=m\dfrac{3v_3^2}{L}$$
The minimum speed occurs when the tension in the wire is reaching zero. Thus,
$$ \color{red}{\bf\not} mg+0= \color{red}{\bf\not} m\dfrac{3v_3^2}{L}$$
Solving for $v_3$;
$$ g = \dfrac{3v_3^2}{L}$$
$$v_3=\sqrt{\dfrac{gL}{3}}\tag 1$$
Now we need to use the conservation of energy law to find the initial height of releasing the ball that will give us this speed of $v_3$ at point 3.
$$E_1=E_3$$
$$U_{1g}+K_1=U_{3g}+K_3$$
we know that the ball is released from rest, so $K_1=0$
$$U_{1g}+0=U_{3g}+K_3$$
$$ \color{red}{\bf\not} mgy_1 = \color{red}{\bf\not} mgy_3+\frac{1}{2} \color{red}{\bf\not} mv_3^2$$
Plugging from (1);
$$ gy_1 = gy_3+\frac{1}{2} \left(\dfrac{gL}{3}\right)$$
From the figure below, we can see that $y_1=h+\dfrac{2L}{3}$ and $y_3=\dfrac{2L}{3}$;
$$ g \left(h+\dfrac{2L}{3}\right)= g \left( \dfrac{2L}{3}\right)+ \dfrac{gL}{6} $$
$$ gh+ \color{red}{\bf\not} \dfrac{2gL}{3} = \color{red}{\bf\not} \dfrac{2gL}{3}+ \dfrac{gL}{6} $$
$$ \color{red}{\bf\not} gh= \dfrac{ \color{red}{\bf\not} gL}{6} $$
$$ h= \dfrac{ L}{6} \tag 2$$
Now we need to find $h$ and as we see in the right-triangle
$$\cos\theta=\dfrac{\dfrac{L}{3}-h}{L}$$
Thus,
$$h=\dfrac{L}{3}-L\cos\theta =L\left( \dfrac{1}{3}-\cos\theta \right)$$
Plugging into (2);
$$ \color{red}{\bf\not} L\left( \dfrac{1}{3}-\cos\theta \right) = \dfrac{ \color{red}{\bf\not} L}{6} $$
$$ \dfrac{1}{3}-\cos\theta = \dfrac{ 1}{6 } $$
$$ \cos\theta =\dfrac{1}{3} -\dfrac{ 1}{6 } =\dfrac{ 1}{6 } $$
Hence,
$$\theta=\cos^{-1}\left[\dfrac{ 1}{6 }\right]=\color{red}{\bf 80.4^\circ}$$
