the set of quantities you offered us to calculate the result is strange, or at least unusual. In particular, there is nothing such as the (dimensionless) absorption probability $P(E)$ for a molecule to absorb a photon.
The absorption probability is given by the cross section which you called $A$. For every process, one has to determine the cross section again. Usually, the cross section is called $\sigma$ instead of $A$.
So the best thing I could do with your numbers and functions would be to imagine that the cross section in your case was given by $\sigma(E) = P(E)\cdot A$. But it is really misleading to factorize the cross section in such ways. In particular, a universal "cross section area $A$" doesn't mean anything. Molecules are not hard balls that have a universal well-defined cross-sectional areas. They're nuclei surrounded by soft wave functions of electrons that reach arbitrarily far but are getting weaker with the distance - in fact, the (one) wave function for $N$ electrons lives in an $3N$-dimensional space rather than the ordinary $3$-dimensional space.
The number of events, in your case absorption events, is simply the product of the integrated luminosity and the cross section $\sigma$. The integrated luminosity is the time integral of luminosity $L$. It is normally written as $L=\rho\times v$ where $\rho$ is the number density of the beam - in particles per unit volume - and $v$ is the velocity - for photons, it's the speed of light $c$. The quantity $L$ happens to be the same thing as your flux $f$. If there are some losses or extra percentages etc., you have to be careful about it.
The quantities such as the cross section are designed in such a way that you are allowed to imagine that the molecule is classical and literally has the cross-sectional area $\sigma$ - that's why they were designed - and you get the right probability. If you do think right, you will indeed find out that only a small percentage of the X-ray photons is absorbed. That's why X-rays are being used to see through human bodies at X-ray pictures.
Best wishes
Lubos
From the Virial Theorem we can say the total energy of the atom is propotional to the potential energy of the atom. The potential energy is given by a coulomb potential and so is it will be roughly proportional to $\frac{1}{\langle r \rangle}$ where $\langle r \rangle$ is the mean radius of the electron's orbital. For a hydrogen atom the energy $E\propto \frac{1}{n^2}$, so we would expect \begin{align}\frac{1}{\langle r \rangle} & \propto \frac{1}{n^2}\\
\langle r \rangle &\propto n^2\end{align}
Unfortunately this does not help you much in storing your infinite amount of information in a single atom. In order to get an estimate of $\langle r \rangle$ you need to make many measurements of the position of the electron (especially if it is in a very spread out distribution such as for a high $n$ state) each of these measurements will collapse the wavefunction and you will have to prepare the atom into its initial state all over again before making the next measurement... but that was exactly what you were trying to ascertain by measuring the electron! To determine the state of your atom you need a set of quantities that can be measured simultaneously and uniquely determine which state your atom was in first time, which for a hydrogen atom essentially means you want the energy and the angular momentum. Now the angular momentum shouldn't give you too much trouble; it is nicely quantised into integer multiples of $\hbar$. The energy however will give you a problem. As $E \propto -n^{-2}$, for large $n$, the separation of energy levels goes like \begin{align} \frac{1}{n^2} - \frac{1}{(n+1)^2} &= \frac{1}{n^2}\left[ 1 - (1+ \frac{1}{n})^{-2}\right]\\ &\approx \frac{2}{n^3}\end{align}
That will get very small very fast and how accurately you can measure that separation will limit how much information you can practically store.
Best Answer
Yes, it does. The wavefunction of the first-excited state is different from the ground state and all the other, higher, excited states. (See Ron Maimon's answer or search the web by "particle in a box".) The difference of the wavefunctions makes difference of properties between the states, such as the probability density of finding the proton at a given position in the box.
For example, if the proton is in the ground state, the probability density to find it at the center of the box is non-zero (actually it is maximum over all the possible position in the box), whereas if the proton is in the first-excited state, the probability density to find it at the center of the box is exactly zero.
As another point, the energy eigenvalue of the first-excited state is different from the ground state and all the other, higher, excited states. The energy of the emitted or absorbed photon depends on the energy difference of the initial and final states of the optical transition. Therefore, whether the proton is in the first-excited state or another higher-excited state makes difference in the photon energy when a photon is emitted by a transition to the ground state.
By the way, given two states, the transition between these two states may be either allowed or prohibited depending on the number of photons emitted/absorbed and depending on the transition is by the dipole/polarizability/hyperpolarizability/.. interaction or quadrupole interaction, etc., with the optical field. A rule whether a transition is allowed or prohibited is generally called a selection rule. This has to do with the symmetries of the interaction as well as the states involved.
In summary, it is not enough to know what system (e.g., a proton in an infinite well) is at hand, but necessary to know what state (e.g., the first-excited state) or a superposition of states (i.e., a wavepacket) the system is in, in order to make any prediction. (Let me try to make a (vague) analogy in classical mechanics. It is not enough to know that a stone of mass 1 kg is flying in a free space, but I would like to know in which direction, at what speed, and at what position it is flying in order to calculate if it is hitting me and to decide if I should call my mama.)