The methyl halides CH₃F and CH₃Cl have dipole moments of 1.85 D and 1.87 D…

Step 1. C-F. $\mu = 1.85 \times 3.336 \times 10^{-30}\ \text{C m} = 6.17 \times 10^{-30}\ \text{C m}$. Then $q = \dfrac{6.17 \times 10^{-30}\ \text{C m}}{1.39 \times 10^{-10}\ \text{m}} = 4.44 \times 10^{-20}\ \text{C}$. As a share of $e$: $\dfrac{4.44 \times 10^{-20}}{1.602 \times 10^{-19}} \times 100 = 27.7\ \%$.

Step 2. C-Cl. $\mu = 1.87 \times 3.336 \times 10^{-30} = 6.24 \times 10^{-30}\ \text{C m}$, so $q = \dfrac{6.24 \times 10^{-30}}{1.78 \times 10^{-10}} = 3.51 \times 10^{-20}\ \text{C}$, which is $21.9\ \%$ of $e$.