If the total energy of an electron in the 1st shell of H atom = 0.0 eV (reference shifted), then its potential energy in the 1st excited state would be

In 1st excited state (n=2): E₂ = −3.4 eV (normal). PE = 2TE = −6.8 eV (normal). Shifting reference so E₁=0 means adding +13.6 eV to all energies. PE(n=2) with shift = −6.8 + 13.6 = +6.8 eV. Answer: +6.8 eV.