Answer
$\alpha_a=\alpha_b\gt \alpha_c=\alpha_d$
Work Step by Step
We know that the net torque is given by
$$\tau=Fr\sin\theta=I\alpha$$
So, the angular acceleration is
$$\alpha=\dfrac{Fr\sin\theta}{I}$$
Recall that the moment of inertia of the given system is given by$I=mr^2$.
Hence,
$$\alpha=\dfrac{F\color{red}{\bf\not} r\sin\theta}{mr^{\color{red}{\bf\not} 2}}$$
$$\alpha=\dfrac{F \sin\theta}{mr}$$
Now we need to find the angular acceleration for each ball.
$$\alpha_a=\dfrac{F_a \sin\theta}{m_ar_a}=\dfrac{F \sin90^\circ}{m r }=\boxed{\dfrac{F }{m r }}$$
$$\alpha_b=\dfrac{F_b \sin\theta}{m_br_b}=\dfrac{\color{red}{\bf\not} 2F \sin90^\circ}{\color{red}{\bf\not} 2m r }=\boxed{\dfrac{F }{m r }}$$
$$\alpha_c=\dfrac{F_c \sin\theta}{m_cr_c}=\dfrac{F \sin90^\circ}{m (2r) }=\boxed{\dfrac{F }{2m r }}$$
$$\alpha_d=\dfrac{F_d \sin\theta}{m_dr_d}=\dfrac{\color{red}{\bf\not} 2F \sin90^\circ}{\color{red}{\bf\not} 2m (2r) }=\boxed{\dfrac{F }{2m r }}$$
Therefore, the rank from the largest to the smallest is as follows;
$$\boxed{\boxed{\alpha_a=\alpha_b\gt \alpha_c=\alpha_d}}$$
