An electron in a hydrogen atom makes a transition from an initial...

The ionization energy of a state is the energy required to remove an electron from that state to infinity ($E = 0$). Thus, the energy of the initial state is $E_i = -0.85 \mathrm{eV}$.

The excitation energy of a state is the energy required to excite the electron from the ground state to that state. Given the ground state energy is $-13.6 \mathrm{eV}$, the energy of the final state is $E_f = -13.6 \mathrm{eV} + 10.2 \mathrm{eV} = -3.4 \mathrm{eV}$.

The energy of the emitted photon during the transition is the difference between the initial and final state energies:
$\Delta E = E_i - E_f = -0.85 \mathrm{eV} - (-3.4 \mathrm{eV}) = 2.55 \mathrm{eV}$.

Answer: $2.55 \mathrm{eV}$