Maximum number of atoms in the ground state I was given the following question.

A sample contains $100$ identical and hypothetical $H$-like atoms ( $Z$ may be fractional). Out of the given atoms, some are in the ground state, while others are in a higher energy level ($n=x$). The $P.E$ of the electrons in the ground state is $-192\ \mathrm{eV}$.
The sample is subject to radiations of wavelength $155\ \mathrm{nm}$, and the electrons jump to a higher energy level ($n=x+3$). Upon back-transition, a total of $10$ different spectral lines are observed
If the final state the electrons reach after absorbing photons of $\lambda=155\ \mathrm{nm}$ is $a$ , and the maximum number of atoms which had electrons in the ground state initially is $b$, find $\frac{b}{a}$

By equating ${}^{x+3}C_2$ to $10$, I found $x$ to be $2$. Then, $a$ must be $x+3$, which is $5$.
As to finding the value of $b$, I am stuck. Can we use the fact that $\frac{|P.E|}{2} = |T.E|$? The energy associated with each photon comes out to be $8\ \mathrm{eV}$, so the total energy absorbed is $800\ \mathrm{eV}$ (Since there are $100$ atoms). Maybe we can use this?
Any help would be greatly appreciated.
Edit:- The answers given are $x+3=6$ and $96$ , so I ended up getting the first part wrong as well.
Of the $100$, initially some are in $n=1$, and some in $n=2$ and then some in $n=5$ after absorbing 155 nm then some in lots of other levels after photons emitted. Thus, eventually, none will be left in $n = 5$ but spread  in all in the other levels  including the ground state. However, all other possible transitions occur ($n=2 \to 1$ for example) so all end up in the ground state in total $(100-b)$. However, you do not know the initial amount $b$ so I'm not sure that this question can be answered except that the ratio is $b/(100-b)$.