Answer
$g\approx 7.5\;\rm m/s^2$
Work Step by Step
First of all, we need to use the kinematic formula of vertical displacement.
$$y_f-y_i=v_{iy}t+\frac{1}{2}a_yt^2$$
The initial velocity is always zero since the ball will be released from rest. Also, the final position $y_f=0$ since it hits the ground each time.
The acceleration here is the free-fall acceleration of planet X.
Thus,
$$ -y_i=-\frac{1}{2}gt^2$$
Let's take the initial height $h$ rather than $y_i$.
$$h= \frac{g}{2}t^2$$
Now we can draw the height as a function of time squared $t^2$.
Recall the equation of the linear line
$$y=mx+b$$
whereas $m$ is the slope of the line and $b$ is the constant that intercepts the $y$-axis. And in our case, $y=h$, $m=\dfrac{g}{2}$, $b=0$, and $x=t^2$
Now we need to plug the given data into a graph in which the time column had to be squared, as we see below.
We have now a linear relation between $h$ and $t^2$.
Now we can find the slope of the best fit line to find the free-fall acceleration of planet X.
From the graph below,
$${\rm Slope}=\dfrac{5.44-0}{1.46-0}=\dfrac{g}{2}=3.73$$
Thus,
$$g=\color{red}{\bf 7.45}\;\rm m/s^2$$
