Two insulating small spheres are rubbed together and then...

When two spheres are rubbed together, they acquire equal and opposite charges. Let the number of electrons transferred be $n$.

The magnitude of charge on each sphere is $q = ne$.

Using Coulomb's law, the force of attraction is given by:

$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r^2}$

Substituting the given values $F = 0.1 \text{ N}$ and $r = 1 \text{ cm} = 10^{-2} \text{ m}$:

$0.1 = \dfrac{9 \times 10^9 \times q^2}{(10^{-2})^2}$

$0.1 = \dfrac{9 \times 10^9 \times q^2}{10^{-4}}$

$q^2 = \dfrac{0.1 \times 10^{-4}}{9 \times 10^9} = \dfrac{10^{-5}}{9 \times 10^9} = \dfrac{1}{9} \times 10^{-14}$

$q = \dfrac{1}{3} \times 10^{-7} \text{ C}$

Since $q = ne$, the number of electrons transferred is:

$n = \dfrac{q}{e} = \dfrac{\dfrac{1}{3} \times 10^{-7}}{1.6 \times 10^{-19}}$

$n = \dfrac{10^{-7}}{4.8 \times 10^{-19}} = \dfrac{10^{12}}{4.8} \approx 2.08 \times 10^{11}$

The approximate number of electrons transferred is $2 \times 10^{11}$.