#### Answer

$\frac{F_e}{F_g}=2.3\times 10^{39}$

#### Work Step by Step

We know that the electric force between electron and proton is given as
$F_e=K\frac{q_1q_2}{r^2}$..........eq(1)
We also know that the gravitational force between proton and electron is given as
$F_g=G\frac{m_1m_2}{r^2}$.....eq(2)
Dividing eq(1) by eq(2), we obtain:
$\frac{F_e}{F_g}=\frac{Kq_1q_2}{Gm_1m_2}$
As charge on electron and proton is same so $q_1=q_2=q$
$\frac{F_e}{F_g}=\frac{Kq^2}{Gm_1m_2}$
We plug in the known values to obtain:
$\frac{F_e}{F_g}=\frac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}(1.67\times 10^{-27})(9.11\times 10^{-31})}$
$\frac{F_e}{F_g}=2.3\times 10^{39}$