Q:

What does it mean for a particle to be prepared in a state with an Energy

For example,
what exactly would it mean, in the case of an infinite quantum well, to prepare a particle with the state $E = 5E_2$
I'm assuming $E_2$ would relate to $E_n = \frac{n^2\hbar^2}{2ma^2} = \frac{k_n^2}{2m}$ where $a$ is the well width.
But what does this tell us about the wave function or the particle and what does it physically mean?

A:

In most situations the statement refers to preparation in an eigenstate of some observable.  This is meant to give information about the quantum state at $t=0$.  Thus, if a system is prepared in a state of energy $E=E_2$, its initial wave function is 
$$
\Psi(x,0)=\psi_2(0)\, .
$$
The difficulty with your example is there is not clearly enough information to reconstruct the initial state.  Since the initial energy of your state is $5E_2=20E_1$, and as there is no eigenstate with this energy, one would have to construct a general combination
$$
\Psi(x,0)=a_1\psi_1(x)+a_2\psi_2(x)+a_3\psi_3(x)+a_4\psi_4(x)
$$
and find the coefficients $a_1,a_2,a_3$ and $a_4$ using only the energy $20E_1$.  The solution is not unique since all you know is that the coefficients must satisfy
\begin{align}
20E_1&=\vert a_1\vert^2 E_1 + \vert a_2\vert^2 4E_1 + \vert a_3\vert^2 9E_1 + \vert a_4\vert^2 16 E_1\, ,\\
1&=\vert a_1\vert^2+\vert a_2\vert^2+\vert a_3\vert^2 +\vert a_4\vert^2
\end{align}
Moreover, as you see, even if you could solve for $\vert a_k\vert^2$, there would remain the question of the relative phase of the terms, and this relative phase has considerable influence on the time-development of the quantum state.

