Answer
$T=\pi\;\sqrt{\dfrac{7L }{ 3g }}$
Work Step by Step
First, we need to find the rod's moment of inertia around the pivot point. So we need to use the parallel-axis theorem.
$$I_{pivot}=I_{cm}+Md^2$$
The pivot point is at a distance of $L/4$ from one end of the rod, and the moment of inertia of a uniform rod around its center of mass is $ML^2/12$.
Thus,
$$I_{pivot}=\frac{1}{12}ML^2+M\left[\dfrac{L}{4}\right]^2$$
$$I_{pivot}=\frac{7}{48}ML^2\tag 1$$
Now we need to find the period 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,
$$2\pi f=\sqrt{\dfrac{Mg\frac{L}{4}}{I}}$$
Plugging from (1);
$$2\pi f=\sqrt{\dfrac{ \color{red}{\bf\not} Mg \color{red}{\bf\not} L}{4(\frac{7}{48} \color{red}{\bf\not} ML^{ \color{red}{\bf\not} 2})}}$$
$$2\pi f=\sqrt{\dfrac{ g }{ \frac{7}{12} L }}$$
Recalling that the period is given by $T=1/f$, so
$$2\pi \dfrac{1}{T}=\sqrt{\dfrac{ g }{ \frac{7}{12} L }}$$
Solving for $T$;
$$T=2\pi\;\sqrt{\dfrac{7L }{ 12g }}$$
$$\boxed{T= \pi\;\sqrt{\dfrac{7L }{ 3g }} }$$