Answer
$0.66\;\rm s$
Work Step by Step
First, we need to find the moment of inertia of the system around the upper end of the rod, and its center of mass as well.
The moment of the inertia of the rod and the clay ball is given by
$$I=I_{\rm end\;rod}+mL^2$$
where $m$ is the mass of the clay ball and $L$ is the length of the rod and the distance from the ball to the axis of rotation.
Recall that the moment of inertia of the rod around one end is $\frac{1}{3}ML^2$.
$$I=\frac{1}{3}ML^2+mL^2\tag 1$$
The center of the mass of the system from the upper end of the rod is given by
$$y_{cm}=\dfrac{M\frac{L}{2}+mL}{M+m}\tag 2$$
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, $l$ is the distance from the pivot point to center of mass of the system.
So, in our case,
$$\omega=2\pi f=\sqrt{\dfrac{(m+M)g(y_{cm})}{I}}$$
where $f=1/T$, so
$$\dfrac{ 2\pi }{T}=\sqrt{\dfrac{(m+M)g(y_{cm})}{I}}$$
$$\dfrac{T}{ 2\pi }=\sqrt{\dfrac{I}{(m+M)g(y_{cm})}}$$
$$ T =2\pi \sqrt{\dfrac{I}{(m+M)g(y_{cm})}}$$
Plugging from (1) and (2);
$$ T =2\pi \sqrt{\dfrac{\frac{1}{3}ML^2+mL^2}{(m+M)g\left(\dfrac{M\frac{L}{2}+mL}{M+m} \right)}}$$
$$ T =2\pi \sqrt{\dfrac{\frac{1}{3}ML^2+mL^2}{ g( M\frac{L}{2}+mL)} }$$
Pluggign the known;
$$ T =2\pi \sqrt{\dfrac{\frac{1}{3}(0.2)(0.15)^2+(0.02)(0.15)^2}{ 9.8( (0.2)\frac{0.15}{2}+(0.02)(0.15))} }$$
$$T=\color{red}{\bf 0.66}\;\rm s$$
