From the first postulate of Bohr's atom model.
$\frac{m v^{2}}{r} =\frac{K Z e^{2}}{r^{2}}$
$\Rightarrow \frac{1}{2} m v^{2} =\frac{1}{2} \frac{K Z e^{2}}{r}$
i.e., $KE$ of electron $=\frac{1}{2} m v^{2}=\frac{K Z e^{2}}{2 r}$ ...(1)
Potential due to the nucleus, in the orbit in which electron is revolving
$=\frac{K Z e}{r}$
$\therefore $ Potential energy of electron
potential $\times$ charge
$=\frac{K Z e}{r}(-e)=-\frac{K Z e^{2}}{r}$ ...(2)
Total energy of electron in the orbit
$E = KE + PE$
$=\frac{1}{2} \frac{K Z e^{2}}{r}-\frac{K Z e^{2}}{r}$
$=-\frac{K Z e^{2}}{2 r}$ ...(3)
This implies that
$PE =2 E$
Given, that in ground state of hydrogen, total energy
$E =-13.6\, eV$
$PE =-2 \times 13.6$
Total energy of electron in outer orbits is more than