Populations of a two level system
I'm getting basically the same answer when I work out the math. I got $T=3398.45\ K$, but we probably just rounded the constants differently.

This is a good lesson in how much thermal energy is carried around at any given time by molecules in the air, as well as the importance of distinguishing between thermal processes and photo-processes.

That is, if we wish to sustain an ensemble of excited particles, there must be sufficient energy such that when we spread the energy out evenly, which is one of the things that this equation implies given that it claims all possible microstates are equally likely, some of the energy must be stored in the form of electronic excitations.

The reason this probably seems like a very high temperature is because we use equipment all the time that relies on electronic excitations and we never have problems with the equipment melting due to it being at thousands of degrees Celsius. That is simply because in these photo-process, meaning the energy is carried in the form of photons, the temperature never is that hot.

That is, if I shoot a photon in, or even a stream of photons at a sample with the transition you are describing, each photon carries a very small amount of energy, and even though there may be quite a lot of them, the energy is quickly dispersed because the excited state is very short-lived, so the photon gets re-emitted, and likely just makes it way out into an infinite reservoir known as the surroundings.

Thus, it is very difficult to maintain a population of excited states thermally, because the system needs to basically be a closed system or else it will eventually just radiate all the energy away.



Another way of thinking about this is that the value of $k_bT$ at room-ish temperature of $298\ K$ (in a familiar unit) is $207\ cm^{-1}$. This makes it clear then why rotationally excited states are frequently occupied at room temperature, because these transitions are on the order of tens of wavenumbers. On the other hand, we often get off free assuming everything is in the vibrational ground state because these transitions are on the order of thousands of wavenumbers.

This is also where the idea of rotational temperature and vibrational temperature come from. These can be understood physically as,


  an estimate of the temperature at which thermal energy (of the order of $k_bT$) is comparable to the spacing between rotational [or vibrational] energy levels.