Answer
$21.2\;\rm rad/s^2$
Work Step by Step
The ball will be released after a $\frac{11}{12}$ revolution of the wheel. The ball's initial position is on the top of the wheel as seen in the given figure and the wheel spins around clockwise.
So, the releasing point will be as seen below.
The releasing velocity is tangential to the wheel as we know.
Thus,
$$v_0=r\omega\tag 1 $$
where $r$ is the radius of the wheel and $\omega$ is the wheel's instantaneous angular velocity.
We know that $\theta$ in the figure below is given by
$$\theta=360^\circ-\left[\frac{11}{12}\cdot360^\circ\right]=\bf 30^\circ$$
Now we can see that the initial height of the ball is
$$y_i=r\cos\theta=0.2 \cos30^\circ$$
while the final height is zero.
and the initial horizontal position is zero while the final position is given by
$$x=r\sin\theta+1.0=0.2\sin30^\circ+1 $$
So, the time of the trip is given by
$$v_x=\dfrac{x}{t}$$
hence,
$$t=\dfrac{x}{v_x}=\dfrac{0.2\sin30^\circ+1}{v_0\cos 30^\circ}$$
$$t=\dfrac{1.1}{v_0\cos30^\circ }\tag 2$$
In the $y$-direction, there is a free-fall acceleration.
$$y=y_i+v_{iy}t-\frac{1}{2}gt^2$$
$$0=0.2 \cos30^\circ+v_{0}\sin30^\circ t-4.9t^2$$
Plugging from (2);
$$0=0.2 \cos30^\circ+v_{0}\sin30^\circ\left[\dfrac{1.1}{v_0\cos30^\circ }\right]-4.9\left[\dfrac{1.1}{v_0\cos30^\circ }\right]^2$$
$$0=0.2 \cos30^\circ+ 1.1\tan30^\circ - \dfrac{4.9\times 1.1^2 }{v_0^2\cos^230^\circ } $$
Thus,
$$v_0=3.127\;\rm m/s\tag 3$$
Solving (1) for $\omega$ and plugging $v_0$;
$$ \omega_f= \dfrac{v_0}{r} $$
The angular acceleration is given by
$$\omega_f^2-\omega_i^2=2\alpha (\theta_f-\theta_i)$$
Thus;
$$\alpha=\dfrac{\omega^2_f-\omega^2_i}{2\theta_f-2\theta_i}=\dfrac{ \left[\dfrac{v_0}{r}\right]^2-0}{2\cdot\frac{11}{12}(2\pi) -0}$$
$$\alpha =\dfrac{ \dfrac{3.127^2}{0.2^2} }{\frac{11}{12}(4\pi) }$$
$$\alpha =\color{red}{\bf 21.2}\;\rm rad/s^2$$