The ionization energy of hydrogen atom is  $1314 \mathrm{kJ} {\mathrm{mol}}^{-1}$. The energy of $n=3$ electronic level in this atom is 

Ionization energy per hydrogen atom is $\mathrm{IE}=\frac{\left(1314\times {10}^3\mathrm{J} {\mathrm{mol}}^{-1}\right)}{\left(6.022\times {10}^{23 }{\mathrm{mol}}^{-1}\right)}=2.182\times {10}^{-18}\mathrm{J}$

The energy of $n=1$ orbit in hydrogen atom will be $E_1=-2.182\times {10}^{-18}\mathrm{J}$

The energy of $n=3$ orbit will be$E_3=\frac{E_1}{3^2}=-\frac{2.182\times {10}^{-18}\mathrm{J}}{9}=-2.42\times {10}^{-19}\mathrm{J}$