Solution

The given ratio is $$\displaystyle \frac{ke^2}{Gm_cm_p}$$

Where,

$$G =$$ Gravitational constant

Its unit is $$N\,m^2kg^{-2}$$.

$$m_e$$ and $$m_p=$$ Masses of electron and proton.

Their unit is kg.

$$e =$$ Electric charge.

Its unit is C.

$$k = A$$ constant

$$\displaystyle =\frac{1}{4\pi\in_0}$$

$$\in_0=$$ Permittivity of free space

Its unit is $$N\,m^2C^{-2}$$.

Unit of the given ratio $$\displaystyle \frac{ke^2}{Gm_cm_p}=\frac{[Nm^2C^2][C^{-2}]}{[Nm^2\,kg^{-2}][kg][kg]}$$

$$=M^0L^0T^0$$

Hence, the given ratio is dimensionless.

$$e=1.6\times 10^{-19}C$$

$$G=6.67\times 10^{-11}N\,m^2kg^{-2}$$

$$m_e=9.1\times 106{-31}kg$$

$$m_p=1.66\times 10^{-27}kg$$

$$\displaystyle \frac{ke^2}{Gm_cm_p} = \frac{9\times 10^9\times(1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-3}\times 1.67\times 10^{-22}}\approx 2.3\times 10^{39}$$

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.