The energy of electron in the first excited state of Hydrogen atom is -3.4 eV. The potential energy of this electron is:

-6.8 eV

The correct answer is Option (3) → -6.8 eV

$E_L$, Energy in first excited state = $\frac{13.6eV}{2^2}=-3.42eV$ [given]

and,

in hydrogen atom, the potential energy is twice the total energy,

$U=2E$

$=2×-3.42eV=-6.8eV$