Answer
$313\;\rm nC$
Work Step by Step
Let's assume that the system [spring+bead+Van de Graaff
generator] is perfectly isolated. Thus the energy is conserved.
Noting that the gravitational effect here is neglected.
$$E_i=E_f$$
The spring was initially compressed which means that the initial energy is the spring's potential energy. At the closest approach, the whole energy will be stored as an electric potential energy.
$$U_{sp}=U_{e}$$
Noting the bead starts from rest and ends at rest as well.
$$\frac{1}{2}kx^2=\dfrac{ q_{\rm _{bead}}\;q_{\rm _{Van}}}{(4\pi \epsilon_0)r}$$
Hence,
$$\dfrac{1}{r}=\left[\dfrac{(4\pi \epsilon_0)k}{2q_{\rm _{bead}}\;q_{\rm _{Van}}}\right]x^2$$
Now we can draw this straight line where $y=1/r$, $x=x^2$, and the slope is $\left[\dfrac{(4\pi \epsilon_0)k}{2q_{\rm _{bead}}\;q_{\rm _{Van}}}\right]$.
So we need to draw the best-fit line and find its slope so we can solve for $q_{\rm _{Van}}$
$${\rm Slope}=\left[\dfrac{(4\pi \epsilon_0)k}{2q_{\rm _{bead}}\;q_{\rm _{Van}}}\right] $$
$$q_{\rm _{Van}}=\left[\dfrac{(4\pi \epsilon_0)k}{2q_{\rm _{bead}}\;{\rm Slope} }\right] \tag 1$$
We should add a 3-cm distance to the closest approach since we had to measure it from the center of the VandeGraaf sphere.
See the graph below.
\begin{array}{|c|c|c|c|}
\hline
x^2& \dfrac{1}{r} \\
\hline
0.016^2&\dfrac{1}{0.085} \\
\hline
0.019^2 &\dfrac{1}{0.056} \\
\hline
0.022^2 & \dfrac{1}{0.046} \\
\hline
0.025^2 & \dfrac{1}{0.034}\\
\hline
\end{array}
Plug these dots and then find the slope of the best-fit line.
Now we need to plug the slope into (1) and the rest of the known;
$$q_{\rm _{Van}}=\left[\dfrac{ 0.65}{2(9\times 10^9)(2.5\times10^{-9})\;(46188) }\right] $$
$$q_{\rm _{Van}}=\color{red}{\bf 3.13\times10^{-7} }\;\rm C=\bf 313\;\rm nC $$
