Solution

a) By using Coulomb's Law:
$$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$
Plugging in the given values ($1\text{ C}$, $2\text{ C}$, $1\text{ km} = 1000\text{ m}$):
$$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{1\cdot 2}{1000^2} = \frac{2}{10^6 \cdot 4\pi\varepsilon_0}$$
Substituting the constant $\frac{1}{4\pi\varepsilon_0} = 9 \cdot 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$, calculating this value gives us:
$$F = 2 \cdot 10^{-6} \cdot 9 \cdot 10^9 = 18,000\text{ N} = 1.8 \cdot 10^4\text{ N}$$

b) Once again, using Coulomb's Law:
$$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$
Knowing that the charge of an electron is $e = -1.6 \cdot 10^{-19}\text{ C}$, and their distance is $10^{-8}\text{ cm} = 10^{-10}\text{ m}$:
$$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{(-1.6 \cdot 10^{-19})^2}{(10^{-10})^2} = 9 \cdot 10^9 \cdot \frac{2.56 \cdot 10^{-38}}{10^{-20}} = 2.3 \cdot 10^{-8}\text{ N}$$
Now we compare this to the gravitational attraction of the electrons, given by the Law of Universal Gravitation:
$$F_g = G\frac{m_1 m_2}{r^2}$$
Plugging in the known distance and the electron mass of $m = 9.1 \cdot 10^{-31}\text{ kg}$:
$$F_g = 6.67 \cdot 10^{-11} \cdot \frac{(9.1 \cdot 10^{-31})^2}{(10^{-10})^2} = 5.5 \cdot 10^{-51}\text{ N}$$
Dividing these two values to find the ratio between electromagnetic and gravitational attraction gives us:
$$\frac{F_e}{F_g} = \frac{2.3 \cdot 10^{-8}}{5.5 \cdot 10^{-51}} = 4.2 \cdot 10^{42}$$