Answer
$\pm 8.2 \;\rm nC $
Work Step by Step
We know that the electric force between the two balls is an attractive force and its magnitude is given by Coulomb’s law.
$$F=\dfrac{kq_1q_2}{r^2}$$
We are given that the magnitude of the two charges in both balls is the same, so $q_1=q_2=q$. And we also know that the separation distance here is $d$.
Thus,
$$F=\dfrac{kq^2}{d^2}\tag 1$$
From Newton's second law, we know that
$$F_{net}=ma\tag 2$$
and since the surface is slippery, the only force that is exerted on one of the balls is the attractive electric force from the other ball.
Hence,
$$F_{net}=F$$
From (1) and (2),
$$\dfrac{kq^2}{d^2}=ma$$
Thus,
$$a=\dfrac{kq^2}{md^2}$$
$$a=\dfrac{kq^2}{m}\cdot \dfrac{1}{d^2} \tag 3$$
where we can assume that this is a straight-line formula $y=mx+b$ where
$y=a$,
$x=\dfrac{1}{d^2}$,
$m={\rm solpe}=\dfrac{kq^2}{m}$,
Hence, we can find the charge from the slope
$${\rm solpe}=\dfrac{kq^2}{m}$$
$$ q=\sqrt{\dfrac{m\;{\rm solpe}}{k}}$$
Plugging the known;
$$ \boxed{q=\sqrt{\dfrac{0.002\;{\rm solpe}}{8.99\times 10^9}}}$$
Now we need to use (3) to find the points to draw the best fit-line and find its slope.
As we see below, the slope is about $3.02\times10^{-4}\;\rm m^3/s^2 $. Plug it into the boxed formula above,
$$ q=\sqrt{\dfrac{0.002\; (3.02\times10^{-4})}{8.99\times 10^9}}=\color{red}{\bf \pm 8.2\times 10^{-9}}\;\rm C $$