Answer
$ \alpha=\dfrac{20\;Tr}{13\;MR^2}$
Work Step by Step
Since the rod’s moment of inertia is negligible, the moment of inertia of the system is the moment of inertia of a sphere at a distance of $\frac{1}{2}R$ from its center.
And we need to find that moment of inertia.
According to the parallel-axis theorem, the moment of inertia of our sphere here is given by
$$I=I_{cm}+Md^2$$
where $d$ here is given by $\frac{1}{2}R$; and the moment o inertia of a sphere around its center of mass is given by $\frac{2}{5}MR^2$
$$I=\frac{2}{5}MR^2+M\left[\dfrac{R}{2}\right]^2$$
$$I=\frac{13}{20}MR^2\tag 1$$
Now we know that the net torque exerted on our system is given by
$$\sum\tau=T\overbrace{R_{rod}}^{=r}=I\alpha$$
Thus, the sphere's angular acceleration is given by
$$\alpha=\dfrac{Tr}{I}$$
Plugging from (1);
$$\boxed{\alpha=\dfrac{20\;Tr}{13\;MR^2}}$$
