Answer
$2.1\times 10^{12}$ electrons.
Work Step by Step
The balls can be treated as point charges. One ball will have charge Q, and the other, -Q. Use Coulomb’s law to relate the magnitude of the charge to the strength of the attractive force.
$$F=\frac{kQ^2}{r^2}$$
$$Q=r\sqrt{\frac{F}{k}}= (0.24m)\sqrt{\frac{0.017N}{8.99\times10^9}}=3.30\times10^{-7}C$$
Find the number of electrons transferred.
$$N=\frac{3.30\times10^{-7}C }{1.60\times10^{-19}C/e}=2.1\times 10^{12}\;electrons$$
