Let total energy of hydrogen like atom at ground state $=-E$
So, 1 st excitation
Energy $=E_{2 n d}-E_{1 s t} $
$ 15 \,eV =\left(\frac{-E}{2^{2}}\right)-\left[\frac{E}{1^{2}}\right]$
$(\because$ First excitation potential is given)
$\Rightarrow \, \frac{3 E}{4}=15 \,eV $
$\Rightarrow \, E=\frac{15 \times 4}{3} eV =20\, eV$
Now, the third excitation energy
$E_{3} =E_{4^{t h}}-E_{1^{s t}} $
$=\left(\frac{-E}{4^{2}}\right)-\left[\frac{E}{1^{2}}\right] $
$E_{3} =+\frac{15 E}{16}$
Now, putting the value of $E$, we get
$E_{3}=\frac{15}{16} \times 20=\frac{75}{4} \,eV$