Answer
$9.74\;\rm m/s^2$
Work Step by Step
We are here dealing with a system that we can consider a small-angle physical pendulum without damping.
Now we need to find the frequency of our physical pendulum and we know that the angular frequency of such one is given by
$$\omega=2\pi f=\sqrt{\dfrac{Mgl}{I}}$$
So,
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{Mgl}{I}}$$
Squaring both sides;
$$f^2=\dfrac{1}{4\pi^2}\dfrac{Mgl}{I}$$
Recall that the moment of inertia of a thin-uniform rod around one end is $\frac{1}{3}ML^2$.
$$f^2=\dfrac{1}{4\pi^2}\dfrac{ \color{red}{\bf\not} Mgl}{ \frac{1}{3} \color{red}{\bf\not} ML^2}$$
Noting that $l$ is the distance from the pivot point to the center of mass of the rod which is $L/2$,
$$f^2=\dfrac{1}{4\pi^2}\dfrac{3g \color{red}{\bf\not} L}{2L^{ \color{red}{\bf\not} 2}}$$
Hence,
$$f^2= \dfrac{3g }{8\pi^2 }\dfrac{1}{L}$$
Now this is a form of a straight line formula where $y=f^2$, $x=\dfrac{1}{L}$, and the slope of this line is given by
$$ {\rm Slope}=\dfrac{3g }{8\pi^2 }$$
Hence,
$$g=\dfrac{ 8\pi^2\;\cdot{\rm Slope}}{3}\tag 1 $$
So we have to draw the best-fit line from the given data and then find its slope then we can find the free-fall acceleration.
\begin{array}{|c|c|c|}
\hline
\dfrac{1}{L}\;({\rm m^{-1}})& f^2\;({\rm Hz^2}) \\
\hline
\hline
\dfrac{1}{1.0}=1 & 0.61^2 \\
\hline
\dfrac{1}{0.8}=1.25& 0.67^2\\
\hline
\dfrac{1}{0.6} =\dfrac{5}{3}& 0.79^2\\
\hline
\dfrac{1}{0.4}=2.5&0.96^2\\
\hline
\end{array}
Plug these data into any software calculator, as we can see below.
From the figure below, we can see that the slope is 0.37, plug that into (1).
$$g=\dfrac{ 8\pi^2\; (0.37)}{3}=\color{red}{\bf 9.74}\;\rm m/s^2 $$
which is a closer answer to $9.8\;\rm m/s^2$ but the difference may appear due to the uncertainty of measurements.
