Let the potential energy of Hydrogen atom in the ground state be zero. Then its orbital energy in the first excited state will be

Total energy in the ground state is

E = PE(U) + KE(K) = 0 + 13.6 eV = 13.6 eV 

Energy required to shift to first excited state is 

${\mathrm{E}}_2-{\mathrm{E}}_1=\mathrm{\Delta E}=\left[-\frac{13.6}{2^2}-\left(-13.6\right)\right]\mathrm{eV}=10.2\mathrm{eV}$
$\therefore$Total energy in the first excited state

  = (13.6 + 10.2) eV = 23.8 eV