Answer
$f=\dfrac{1}{2\pi }\sqrt{\dfrac{5g}{7R}}$
Work Step by Step
First, we need to find the moment of inertia of the sphere around the pivot point.
$$I_{pivot}=I_{cm}+Md^2$$
where $d=R$, and the moment of inertia of the sphere is $\frac{2}{5}MR^2$.
$$I_{pivot}=\frac{2}{5}MR^2+MR^2$$
$$I_{pivot}=\frac{7}{5}MR^2\tag 1$$
Now we need to find the frequency of our physical pendulum and we know that the angular frequency of such one is given by
$$\omega=2\pi f=\sqrt{\dfrac{mgl}{I}}$$
where $m$ is the mass of the system, and $l$ is the distance from the pivot point to the center of mass of the system.
So, in our case,
$$ 2\pi f=\sqrt{\dfrac{MgR}{I}}$$
Plugging from (1);
$$ 2\pi f=\sqrt{\dfrac{ \color{red}{\bf\not} Mg \color{red}{\bf\not} R}{\frac{7}{5} \color{red}{\bf\not} MR^{ \color{red}{\bf\not} 2}}}$$
$$ \boxed{f=\dfrac{1}{2\pi }\sqrt{\dfrac{5g}{7R}}}$$
