Answer
$64.6\;\rm g$
Work Step by Step
We need to find an equation that gives us a straight line that has a slope of $m$ (the mass of the ball).
We choose our system to be the spring plus the ball plus the Earth. This system is an isolated system.
Thus, the energy is conserved. So, the energy of the system just before releasing the ball is equal to its energy when the ball reaches the highest point.
$$E_i=E_f$$
Thus,
$$U_{i,sp}+\overbrace{U_{ig,ball}}^{0}+\overbrace{K_{i,ball}}^{0}=\overbrace{U_{f,sp}}^{0}+U_{fg,ball}+\overbrace{K_{f,ball}}^{0}$$
where the ball's initial kinetic and final kinetic energy are zero since its speed is zero at these two positions.
The final elastic potential energy of the spring is zeros since it is now back to its normal length after firing the ball.
We chose the releasing point when the spring is compressed to be our origin, so the intial gravitational poterinal energy is also zero.
$$U_{i,sp}= U_{fg,ball}$$
Thus,
$$\frac{1}{2}ks^2= mgy $$
where $x$ is the compressed distance and $s$ is the highest point the ball reaches.
Solving for $y$;
$$y=\frac{k}{2mg}\;s^2 $$
Now we will treat this equation as a straight line where $s^2$ here is $x$ in $y=mx+b$ [the straight line equation].
The slope of this line is
$${\rm Slope}=\dfrac{k}{2mg}$$
So that the mass of the ball is given by
$$m=\dfrac{k}{2 g\;{\rm Slope}}\tag1$$
\begin{array}{|c|c| }
\hline
y\;({\rm m})& s^2\;({\rm m^2}) \\
\hline
0.32 &4\times 10^{-4} \\
\hline
0.65 & 9\times 10^{-4} \\
\hline
1.15 & 1.6\times 10^{-3} \\
\hline
1.89 & 2.5\times 10^{-3} \\
\hline
\end{array}
As we see in the graph below, we ploted the dots and connected them with the best fit line.
The software calculator shows that the slope of this line is as seen in the figure below, so we need to plug that into (1).
$$m=\dfrac{k}{2 g\;{\rm Slope}} =\dfrac{950}{2\times 9.8\times 750}=\bf 0.0646\;\rm kg$$
$$m=\color{red}{\bf 64.6}\;\rm g$$
