Answer
$8.42\;\rm m/s^2$
Work Step by Step
Let's assume that the pulley is frictionless. So the tension in the string is equal to the mass of the hanging 4-kg block.
$$T_s=mg'$$
$$T_s=4g'\tag 1$$
where $g'$ is the free-fall acceleration on planet X.
The length of the vibrated part of the string is only 2 m ( not 2.5 m) since it is fixed by the pulley-block system at 2 m away.
But the linear density of the string is then given by the full length of the string,
$$\mu=\dfrac{m_{string}}{L_{string}}\tag 2$$
We know that the frequency of the standing waves in a string is given by
$$f_m=\dfrac{mv}{2L}$$
where $L$ is the vibrated part of the string, and $v=\sqrt{T_s/\mu}$
Thus,
$$f_m=\dfrac{m }{2L}\sqrt{\dfrac{T_s}{\mu}}$$
Plugging from (1) and (2);
$$f_m=\dfrac{m }{2L}\sqrt{\dfrac{4g'L_{string}}{m_{string}}}$$
Therefore, squaring both sides.
$$f_m^2=\left[\dfrac{m^2 }{4L^2} \dfrac{4L_{string}}{m_{string}}\right] g'$$
$$f_m^2=\left[\dfrac{m^2 }{ L^2} \dfrac{ L_{string}}{m_{string}}\right] g'$$
Plugging the known;
$$f_m^2=\left[\dfrac{m^2 }{ 2^2} \cdot\dfrac{ 2.5}{0.005}\right] g'$$
$$\boxed{f_m^2= (125g')m^2 }$$
This can be represented as a straight line equation $y=ax+b$ where $y=f_m^2$, $x=m^2$, $a={\rm slope}=125g'$, and $b=0$.
Plugging the data into a table as seen below then plug the dots to draw the best fit line.
\begin{array}{|c|c|c|c|}
\hline
x= m^2& y=f_m^2\;\rm Hz^2 \\
\hline
1& 961 \\
\hline
4 &4356\\
\hline
9 &9025\\
\hline
16 &16900\\
\hline
25 &26244\\
\hline
\end{array}
$${\rm Slope}=125g'$$
Thus,
$$g'=\dfrac{{\rm Slope}}{125}=\dfrac{1052.84}{125}$$
$$g'=\color{red}{\bf 8.42}\;\rm m/s^2$$
