Answer
$2.55\;eV$
Work Step by Step
A hydrogen atom is in a state having a binding energy (the energy required to remove an electron) of $0.85\;eV$.
$E_1=-0.85\;eV$
The atom makes a transition to a state (having the binding energy say $E_2$) with an excitation energy of $10.2\;eV$.
This $10.2\;eV$ energy is actually the energy difference between the energy of the state and that of the ground state
$E_2-E_0=10.2\;eV\\
E_2-(-13.6\;eV)=10.2\;eV\\
E_2=-3.4\;eV$
Now,
$E_1-E_2=E_p\\
or, E_p=[-0.85-(-3.4)]\;eV\\
or, E_p=2.55\;eV$
Therefore, the energy of the photon emitted as a result of the transition is $2.55\;eV$
