#### Answer

$q = 33.2~nC$

#### Work Step by Step

We can find the spring constant $K$:
$Kx = mg$
$K = \frac{mg}{x}$
$K = \frac{(0.0010~kg)(9.8~m/s^2)}{0.010~m}$
$K = 0.98~N/m$
We can find the magnitude of the charge on each bead:
$F = \frac{kq^2}{r^2} = Kx$
$q^2 = \frac{Kx~r^2}{k}$
$q = \sqrt{\frac{Kx~r^2}{k}}$
$q = \sqrt{\frac{(0.98~N/m)(0.0050~m)~(0.045~m)^2}{9.0\times 10^9~N~m^2/C^2}}$
$q = 33.2\times 10^{-9}~C$
$q = 33.2~nC$