= Solution
The enlarged diamonds reach $|V_{sd}|=5.75\,\mathrm{mV}$ instead of $5.00\,\mathrm{mV}$. Subtract the unchanged charging contribution:
$$
\Delta\epsilon=e(5.75-5.00)\,\mathrm{mV}
=0.75\,\mathrm{meV}.
$$
Thus the two observed closed-orbital gaps are each approximately $0.75\,\mathrm{meV}$.
Define $A=h^2/(8m^*L_x^2)$ and $B=h^2/(8m^*L_y^2)$, and choose the labels so $A\leq B$. With two spins per orbital, the first two closed orbitals correspond to occupancies $N=2$ and $N=4$. The lowest state is $(1,1)$ and the next is $(2,1)$, so
$$
E_{21}-E_{11}=3A=0.75\,\mathrm{meV},
\qquad A=0.25\,\mathrm{meV}.
$$
The next orbital is $(1,2)$, yielding the second observed gap
$$
E_{12}-E_{21}=3B-3A=0.75\,\mathrm{meV},
\qquad B=0.50\,\mathrm{meV}.
$$
This ordering is self-consistent: $E_{11}=0.75$, $E_{21}=1.50$, $E_{12}=2.25$ and $E_{31}=2.75\,\mathrm{meV}$, as shown in the original level diagram.
Using $m^*=0.6m_e$ gives
$$
L_x=\frac h{\sqrt{8m^*A}}\simeq50.2\,\mathrm{nm},
\qquad
L_y=\frac h{\sqrt{8m^*B}}\simeq35.5\,\mathrm{nm}.
$$
The axes can be interchanged. Assigning the same $0.75\,\mathrm{meV}$ directly to both $3A$ and $3B$ would instead give a square box with degenerate $(2,1)$ and $(1,2)$ orbitals. That would require four electrons in that shell and would not give the stated second closed-orbital occupancy $N=4$.
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