Answer
The charge on the ball is $~~-2.5\times 10^{-8}~C$
Work Step by Step
We can find the magnitude of the charge on the ball:
$F = \frac{k~q_b~q_s}{r^2}$
$q_b = \frac{F~r^2}{k~q_s}$
$q_b = \frac{(9.2\times 10^{-4}~N)(0.625~m)^2}{(9.0\times 10^9~N~m^2/C^2)~(1.0\times 10^{13})(1.6\times 10^{-19}~C)}$
$q_b = 2.5\times 10^{-8}~C$
Since the force on the ball is directed away from the sphere, the charge on the ball must be negative.
The charge on the ball is $~~-2.5\times 10^{-8}~C$
