\(E=\frac{-13.6}{n^{2}} Z^{2}\)
\(E_{4 t h}=-13.6 \times \frac{1}{4^{2}}\)
\(=-0.85 eV\)
The energy released is equal to the total energy minus ionization energy of fourth orbit,
\(E_{\text {release}}=E-E_{\text {4th}}\)
\(=15-0.85\)
\(=14.15 eV\)