= Solution
Inside the rectangle the effective-mass <Schrödinger equation> is
$$
-\frac{\hbar^2}{2m^*}\left(\partial_x^2+\partial_y^2\right)\psi=E\psi,
$$
with $\psi=0$ at all four walls because the confining potential is infinite outside. Separate $\psi(x,y)=X(x)Y(y)$. Each factor obeys a one-dimensional <particle in a box> equation. For $x$,
$$
X''+k_x^2X=0,\qquad
X(0)=X(L_x)=0,
$$
so $X$ is proportional to $\sin(n_x\pi x/L_x)$ with $n_x=1,2,\ldots$; the zero and negative integers give no new physical states. The same argument applies in $y$.
The normalized spatial states and their energies are
$$
\psi_{n_x,n_y}(x,y)=\frac2{\sqrt{L_xL_y}}
\sin\left(\frac{n_x\pi x}{L_x}\right)
\sin\left(\frac{n_y\pi y}{L_y}\right),
$$
$$
E_{n_x,n_y}=E_x+E_y
=\frac{h^2}{8m^*}
\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right).
$$
The product form and energy sum follow from separability. Each spatial orbital can accommodate two opposite-spin electrons under the stated spin degeneracy. These are single-particle confinement energies; the electrostatic charging contribution must still be added when interpreting transport addition energies.
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