A hydrogen atom in a state with a binding energy of 0.85 eV...

The binding energy of an electron in the $n^{\text{th}}$ state of a hydrogen atom is given by $B.E. = \dfrac{13.6}{n^2} \text{ eV}$.

For the initial state, $B.E. = 0.85 \text{ eV}$:

$\dfrac{13.6}{n_i^2} = 0.85 \Rightarrow n_i^2 = 16 \Rightarrow n_i = 4$

The excitation energy of a state is the energy required to excite an electron from the ground state ($n=1$) to that state. For the final state, the excitation energy is $10.2 \text{ eV}$:

$E_{n_f} - E_1 = 10.2 \text{ eV}$

$-\dfrac{13.6}{n_f^2} - (-13.6) = 10.2$

$\dfrac{13.6}{n_f^2} = 13.6 - 10.2 = 3.4 \Rightarrow n_f^2 = 4 \Rightarrow n_f = 2$

The energy of the emitted photon during the transition from $n_i = 4$ to $n_f = 2$ is:

$\Delta E = E_4 - E_2 = -0.85 - (-3.4) = 2.55 \text{ eV}$

The wavelength of the emitted radiation is:

$\lambda = \dfrac{hc}{\Delta E} = \dfrac{1242 \text{ eV nm}}{2.55 \text{ eV}} \approx 487 \text{ nm}$