Answer
$-4.22014\times10^{-97}\;\rm J$
$1.20264\times10^{29}\;\rm m$
Work Step by Step
We can use the same formula of the electric force of Bohr orbits with some editing.
We know, for the first Bohr orbit, that
$$E_1=\dfrac{-2\pi^2Z^2e^4m_ek^2}{h^2}$$
To find the energy if the gravitational force is applied, we need to replace $kZe^2$ with $Gm_em_p$ where $G$ is the universal gravitational constant, $m_e$ is the mass of the electron, and $m_p$ is the mass of the proton.
Thus,
$$E_1=\dfrac{-2\pi^2m_e( k^2Z^2e^4)}{h^2}=\dfrac{-2\pi^2m_e( k Z e^2)^2}{h^2}$$
Replacing;
$$E_1= \dfrac{-2\pi^2m_e( Gm_em_p)^2}{h^2}$$
$$E_1= \dfrac{-2\pi^2G^2m_e^3 m_p^2}{h^2}$$
Plugging the known;
$$E_1= \dfrac{-2\pi^2(6.672\times10^{-11})^2\cdot (9.11\times10^{-31})^3 \cdot (1.67\times10^{-27})^2}{(6.626\times10^{-34})^2}$$
$$E_1=\color{red}{\bf -4.22014\times10^{-97}}\;\rm J$$
Now we need to find the radius, of the first Bohr orbit, by the same approach;
$$r_1 =\dfrac{h^2}{4\pi^2 m_e (kZe^2)}$$
$$r_1 =\dfrac{h^2}{4\pi^2 m_e (Gm_em_p)}$$
$$r_1 =\dfrac{h^2}{4\pi^2G m_e^2 m_p }$$
Plugging the known;
$$r_1 =\dfrac{(6.626\times10^{-34})^2}{4\pi^2 (6.672\times10^{-11}) (9.11\times10^{-31})^2 (1.67\times10^{-27})}$$
$$r_1=\color{red}{\bf 1.20\times10^{29}}\;\rm m$$
which is much greater than the distance between the Earth and the sun!
It is even larger than the diameter of our Milky Way galaxy by 120 million times. That's why the electric forces are more significant here.