Solution

(a) By using Coulomb's Law,
\begin{equation} F = \frac{1}{4\pi\epsilon_0}\cdot\frac{q_{1}q_{2}}{r^2} \end{equation}
And plugging in the values given (1 C, 2C, 1 km/1000m),
\begin{equation} F = \frac{1}{4\pi\epsilon_0}\cdot\frac{1\cdot2}{1000^2} = \frac{2}{10^{6}4\pi\epsilon_0} \end{equation}
Calculating this value gives us:
\text{18,000 N or about $1.8\cdot 10^4$ N}.

(b) Once again, using Coulomb's Law:
\begin{equation} F = \frac{1}{4\pi\epsilon_0} \cdot\frac{q_{1}q_{2}}{r^2} \end{equation}
Knowing that the charge of an electron is $-1.6\cdot10^{-19}$, and their distance is $10^{-8}$ cm $= 10^{-10}$ m,
\begin{equation} F = \frac{1}{4\pi\epsilon_0}\cdot\frac{({-1.6\cdot10^{-19}})^2}{({10^{-10}})^2} = \frac{2.56\cdot10^{-38}}{10^{-20}4\pi\epsilon_0} \end{equation}
Computing this value gives $2.3\cdot10^{-8}$ N,
if we compare this to the gravitational attraction of the electrons, which is first given by the Law of Universal Gravitation:
\begin{equation} F = G\frac{m_{1}m_{2}}{r^2} \end{equation}
and then plugging in the known distance and the electron mass of $9.1\cdot10^{-31}$ kg,
\begin{equation} F = G\frac{({-9.1\cdot10^{-31}})^2}{({10^{-10}})^2} = 5.5\cdot10^{-51} \text{ N} \end{equation}
Dividing these two values to give us the ratio between electromagnetic and gravitational attraction gives us:
\begin{equation} \frac{2.3\cdot10^{-8}}{5.5\cdot10^{-51}} = 4.2\cdot10^{42} \end{equation}