= Solution
The interior <time-independent Schrödinger equation> for a free particle is
$$
\boxed{-\frac{\hbar^2}{2m}(\psi_{xx}+\psi_{yy})=E\psi.}
$$
Confinement by the rectangular infinite well imposes <Dirichlet boundary conditions>: $\psi=0$ at $x=0,b$ and $y=0,\sqrt5b$. The wave function is <square-integrable> and may be normalized to unit <probability> on the rectangle.
Separation of variables gives products of the one-dimensional well states,
$$
\psi_{n,\ell}(x,y)=\frac{2}{5^{1/4}b}\sin\frac{n\pi x}{b}\sin\frac{\ell\pi y}{\sqrt5b},
\qquad n,\ell=1,2,\ldots.
$$
The Hamiltonian is the sum of the independent $x$ and $y$ kinetic-energy operators, so the one-dimensional energies add. Put $C=\pi^2\hbar^2/(2mb^2)$. Then
$$
E_{n,\ell}=C\left(n^2+\frac{\ell^2}{5}\right).
$$
For $n=1$, increasing $\ell$ gives $6C/5,9C/5,14C/5,21C/5$. The smallest energy with $n\geq2$ is also $21C/5$, attained at $(2,1)$; all $n\geq3$ give larger energies. Thus the ground and first three distinct excited energy levels are
$$
\boxed{\begin{array}{c|c|c}
\text{level}&E& (n,\ell)\\\hline
\text{ground}&6C/5&(1,1)\\
\text{first excited}&9C/5&(1,2)\\
\text{second excited}&14C/5&(1,3)\\
\text{third excited}&21C/5&(1,4),(2,1)
\end{array}}
$$
The largest listed energy therefore has <energy-level degeneracy> $\boxed{2}$, since the two product eigenfunctions are linearly independent and no other pair solves $5n^2+\ell^2=21$.
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