A hydrogen atom initially in the n = 6 state undergoes two...

Energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\dfrac{13.6}{n^2} \text{ eV}$.

For the initial state $n = 6$, the energy is $E_6 = -\dfrac{13.6}{6^2} = -0.377 \text{ eV}$.

During the first transition, a photon of energy $1.13 \text{ eV}$ is emitted. The energy of the intermediate state $n'$ is $E_{n'} = E_6 - 1.13 \text{ eV}$.

$E_{n'} = -0.377 - 1.13 = -1.507 \text{ eV}$.

Equating this to the energy formula for the $n'$-th state:

$-\dfrac{13.6}{n'^2} = -1.507$

$n'^2 = \dfrac{13.6}{1.507} \approx 9$

$n' = 3$

The intermediate state is $n = 3$.