A molecule of hydrogen chloride has a dipole moment of 3.6 × 10⁻³⁰ C m…

Step 1. On the axis. Use SI units for every quantity: $V = \dfrac{kp}{r^2} = \dfrac{(9.0 \times 10^{9})(3.6 \times 10^{-30})}{(1.2 \times 10^{-9})^2}\ \text{V} = 0.0225\ \text{V}$, that is $22.5$ mV.

Step 2. At $60^\circ$. $V = \dfrac{kp\cos 60^\circ}{r^2} = 0.5 \times 22.5\ \text{mV} = 11.25\ \text{mV}$.

Step 3. The charge. $p = ql$, so $q = \dfrac{p}{l} = \dfrac{3.6 \times 10^{-30}\ \text{C m}}{1.27 \times 10^{-10}\ \text{m}} = 2.83 \times 10^{-20}\ \text{C}$.

Step 4. As a fraction of $e$. $\dfrac{2.83 \times 10^{-20}\ \text{C}}{1.6 \times 10^{-19}\ \text{C}} = 0.177$.