Answer
$\approx 67^\circ$
Work Step by Step
We can assume that the angular momentum is conserved since the rod is hanging from a frictionless horizontal axle.
Thus, the angular momentum around the axle;
$$L_1=L_2 $$
where $_1$ just before the collision and $_2$ is just after the collision.
$$m_bv_{1 }R= I_{sys}\omega_2 $$
where $m_b$ is the mass of the ball.
and $R=\dfrac{L}{2}$
$$m_bv_{1 }\dfrac{L}{2}= I_{sys}\omega_2 \tag 1$$
To find the moment of inertia of the system, we can use the parallel-axis theorem.
$$I_{sys}=I_{cm,rod}+m_bd^2$$
where $I_{cm,rod}=\frac{1}{12}ML^2$
$$I_{sys}=\frac{1}{12}ML^2+m_b\left[\dfrac{L}{2}\right]^2\tag 2$$
Plugging into (1);
$$m_bv_{1 }\dfrac{L}{2}= \left(\frac{1}{12}ML^2+m_b \dfrac{L^2}{4} \right)\omega_2 $$
We can see that $\omega_2=\dfrac{v_2}{R}=\dfrac{v_2}{\frac{L}{2}}=\dfrac{2v_2}{L}$.
Hence,
$$m_bv_{1 }\dfrac{ \color{red}{\bf\not}L}{2}=\dfrac{ \color{red}{\bf\not}L^{ \color{red}{\bf\not}2}}{4} \left(\frac{1}{3}M +m_b \right)\dfrac{2v_2}{ \color{red}{\bf\not}L}$$
$$m_bv_{1 }= v_2\left(\frac{1}{3}M +m_b \right)\tag 3 $$
After the collision, the energy is conserved as well.
$$E_2=E_3 $$
$$\frac{1}{2}I_{sys}\omega_2^2=\frac{1}{2}I_{sys}\overbrace{\omega_3^2}^{=0}+m_b gh$$
the final angular velocity of the system at the top point is zero.
Noting that the clay ball is the only object that moves upward and its gravitational potential energy increases while the gravitational potential energy of the rod does not change since its center of mass is stationary.
$$\frac{1}{2}I_{sys}\omega_2^2=0 +m_b gh$$
Plugging from (2);
$$\frac{1}{2}\left(\frac{1}{12}ML^2+m_b \dfrac{L^2}{4} \right)\omega_2^2= m_b gh$$
$$\frac{1}{2}\dfrac{L^2}{4}\left(\frac{1}{3}M +m_b \right) \left( \dfrac{2v_2}{ L}\right)^2= m_b gh$$
$$\frac{1}{2}\dfrac{ \color{red}{\bf\not}L^2}{ \color{red}{\bf\not}4}\left(\frac{1}{3}M +m_b \right) \left( \dfrac{ \color{red}{\bf\not}4v_2^2}{ \color{red}{\bf\not}L^2}\right)= m_b gh$$
$$\dfrac{1}{2}\left(\frac{1}{3}M +m_b \right) v_2^2 = m_b gh\tag 4$$
From the geometry of the last figure below, we can see that $h$ is given by
$$\cos\theta=\dfrac{\dfrac{L}{2}-h}{\dfrac{L}{2}}$$
$$h = \dfrac{L}{2}-\dfrac{L}{2}\cos\theta = \dfrac{L}{2}\left[1 - \cos\theta \right]$$
Plugging into (4);
$$\dfrac{1}{ \color{red}{\bf\not}2}\left(\frac{1}{3}M +m_b \right) v_2^2 = m_b g \dfrac{L}{ \color{red}{\bf\not}2}\left[1 - \cos\theta \right]$$
$$ \left(\frac{1}{3}M +m_b \right) v_2^2 = m_b gL \left[1 - \cos\theta \right]$$
Plugging $v_2$ from (3);
$$ \left(\frac{1}{3}M +m_b \right) \left[\dfrac{m_bv_1}{\left(\frac{1}{3}M +m_b \right)}\right]^2 = m_b gL \left[1 - \cos\theta \right]$$
$$ \color{red}{\bf\not}\left(\frac{1}{3}M +m_b \right) \dfrac{m_b^{ \color{red}{\bf\not}2}v_1^2}{\left(\frac{1}{3}M +m_b \right)^{ \color{red}{\bf\not}2}} = \color{red}{\bf\not}m_b gL \left[1 - \cos\theta \right]$$
$$ \dfrac{m_b v_1^2}{\left(\frac{1}{3}M +m_b \right) } = gL \left[1 - \cos\theta \right]$$
$$ \dfrac{m_b v_1^2}{ gL \left(\frac{1}{3}M +m_b \right) } = 1 - \cos\theta $$
Hence,
$$ \cos\theta = 1 - \dfrac{m_b v_1^2}{ gL \left(\frac{1}{3}M +m_b \right) } $$
$$ \theta = \cos^{-1}\left[ 1 - \dfrac{m_b v_1^2}{ gL \left(\frac{1}{3}M +m_b \right) } \right]$$
Plugging the known;
$$ \theta = \cos^{-1}\left[ 1 - \dfrac{(10\times10^{-3})(2.5)^2}{ (9.8)(0.30)\left(\frac{1}{3}(75\times10^{-3}) +(10\times10^{-3})\right) } \right]$$
$$\theta_{max}=\color{red}{\bf 66.9}^\circ$$
