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

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

For the upper state, the binding energy is $0.85 \text{ eV}$.

$\dfrac{13.6}{n_2^2} = 0.85 \Rightarrow n_2^2 = \dfrac{13.6}{0.85} = 16 \Rightarrow n_2 = 4$

The excitation energy of a state $n$ is the energy required to excite the electron from the ground state ($n=1$) to that state.

$\Delta E = 13.6 \left(1 - \dfrac{1}{n_1^2}\right) \text{ eV}$

For the lower state, the excitation energy is $10.2 \text{ eV}$.

$13.6 \left(1 - \dfrac{1}{n_1^2}\right) = 10.2$

$1 - \dfrac{1}{n_1^2} = \dfrac{10.2}{13.6} = 0.75$

$\dfrac{1}{n_1^2} = 0.25 \Rightarrow n_1^2 = 4 \Rightarrow n_1 = 2$

The quantum numbers of the upper and lower energy states are $4$ and $2$.