I conjecture that atomic nucleus with $N\approx 250$ particles could be interpreted as a system in equilibrium (maybe the number is too low to consider the use of statistical mechanics as a reasonable approximation). As such I understand that it could be assigned a well-defined temperature.
My question is: What temperature would the atomic nucleus, given the average kinetic energy of the particles that form it?
A nucleus is a quantum system and has states in the same way an atom has states, though the states of nuclei are poorly understood compared to the states of atoms. We can assign a temperature to the nucleus using the Boltzmann distribution i.e. we expect the probability that the nucleus is in a state of energy $E$ to be given by:
$$ P(E) = e^{-E/kT} $$
where $T$ is the temperature of the nucleus.
The trouble is that the energy states of nuclei are generally separated by energies of a MeV or even tens of MeV, while the thermal energy at room temperature is about one fortieth of an electron volt. That means the probability of even the lowest energy state being occupied is around $e^{-40000000}$, which is effectively zero. That means as far as the nucleus is concerned the temperature is always indistinguishable from absolute zero.
You mention:
the average kinetic energy of the particles that form it?
but the nucleus is not like some collection of little balls all orbiting around each other. The protons and neutrons are delocalised and behave more like fuzzy clouds than the particles in an ideal gas. Thus there is no kinetic energy that is related to temperature in the way that the kinetic energy of gas particles depends on the temperature. All we have are the excited states of the nucleus, and these are widely spaced because the nuclear force is so strong.
“Temperature” is the differential relationship between entropy and internal energy,
\begin{align} dU &= T\,dS + \cdots \\ \frac{\partial S}{\partial U} &= \frac 1T \end{align}
where the entropy is the logarithm of the number of states available to the system,
$$ S =k\ln\Omega $$
A nucleus in a state with spin $J$ has multiplicity $\Omega=J(J+1)$. The amount of energy you can add to a nucleus is quantized, with the first excited state for most nuclei around a mega-eV. If you try to add energy to a nucleus in lumps less than a mega-eV, what happens to the entropy is … nothing. You cannot increase the multiplicity of states available to the system if you can’t acccess the first excited state. So in that sense, the temperature is not well-defined.
You could also say that a nucleus at room temperature has zero heat capacity, since it cannot exchange energy with its environment. Zero heat capacity is a zero-temperature phenomenon. A nucleus in thermal equilibrium with a room-temperature thermal bath is quantum-mechanically indistinguishable from a nucleus in thermal equilibrium with a zero-temperature thermal bath.
Note that a nucleus with nonzero spin may have different orientational energy states in a magnetic field. People who make polarized nuclei talk about their “spin temperature.”
Other answers correctly point out that since the energy-level spacing of a nucleus is large (of order MeV or tens of MeV), then unless the nucleus is in a very unusual environment, its state will either not be thermal (e.g. because it got excited by an incident cosmic ray or something) or else its state will be practically indistinguishable from zero temperature so we may as well say it is zero temperature.
But if one wanted to insist that it can't be exactly zero, then, fair enough, it is not quite zero. But if we are saying it is a thermal equilibrium state, then it is going to have to be one in thermal equilibrium with the surrounding electrons. If they are in thermal equilibrium then their temperature will be equal to that of their surroundings if they are part of a lump of condensed matter (liquid or solid). So we end up concluding that the temperature of the nucleus will be room temperature, for a nucleus in an atom in a condensed material at room temperature.
The above is based on standard reasoning about increase of entropy during relaxation to equilibrium; the subsystem with the larger $\partial U/\partial S$ tends to give energy to the one with smaller $\partial U/\partial S$.
I think this answer is correct, but I will add that I am not as sure of it as I would like, so I will be happy to be corrected!
All the other answers consider nuclei at room temperature, and point out that the energy separation scales are such that for almost all (but not all) nuclei, room temperature is essentially equivalent to absolute zero.
But nuclei on Earth aren't always near room temperature: in particle accelerators, we can heat atomic nuclei up much much hotter - up to 5-6 trillion degrees Kelvin. Remarkably, these "quark-gluon plasmas" equilibrate fast enough to reach a well-defined temperature even during their brief existence. In these exotic conditions - and those created by much smaller particle accelerators as well - excited nuclear states definitely come into play and are are strongly populated.
