(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} \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}