= Solution
Let
$$
u_n(x)=\sqrt{\frac2a}\sin\left[\frac{n\pi(x+a/2)}a\right],\quad n=1,2,\ldots,
\qquad \varepsilon=\frac{\pi^2\hbar^2}{2ma^2},\quad J=\lambda\hbar^2.
$$
The normalized orbital <energy eigenstates> vanish outside the well and have one-particle energies $n^2\varepsilon$. Define the normalized symmetric and antisymmetric combinations for $r\ne s$ by
$$
\Phi_{rs}^\pm=\frac{u_r(x_1)u_s(x_2)\pm u_s(x_1)u_r(x_2)}{\sqrt2},
\qquad\Phi_{rr}^+=u_r(x_1)u_r(x_2).
$$
The normalized spin functions are the <singlet state> $\chi_{00}=(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt2$ and the <triplet states> $\chi_{11}=|\uparrow\uparrow\rangle$, $\chi_{10}=(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)/\sqrt2$, $\chi_{1,-1}=|\downarrow\downarrow\rangle$. Their exchange energies are respectively $3J/4$ and $-J/4$.
For $0<J\ll\varepsilon$, the three lowest distinct <energy levels> of the <exchange-coupled spin-one-half particles in an infinite square well> are
$$
\boxed{\begin{array}{c|c|c}
\text{energy}&\text{multiplicity}&\text{normalized eigenfunctions}\\ \hline
2\varepsilon+3J/4&1&\Phi_{11}^+\chi_{00}\\
5\varepsilon-J/4&3&\Phi_{12}^-\chi_{1M},\ M=-1,0,1\\
5\varepsilon+3J/4&1&\Phi_{12}^+\chi_{00}
\end{array}}
$$
The spatially doubly occupied orbital can only occur with the <singlet state>. The next orbital sum is $1^2+2^2=5$, where either exchange symmetry is possible. The next unused sum is $2^2+2^2=8$, well separated under the small-coupling assumption. The paper's inequality is equivalent, up to a numerical factor, to $|J|\ll\varepsilon$; the factor $\hbar^2$ in $J$ is essential because the <spin> operators carry units of angular momentum.
The sign of $\lambda$ is not explicitly specified. The table assumes the usual intended positive coupling in the displayed minus-sign convention. For $J<0$ with $|J|\ll\varepsilon$, the same three sets of <eigenfunctions> occur, but the two rows at $5\varepsilon$ exchange order. At $J=0$ those rows merge into a single <energy level> of multiplicity four; if three distinct energies are requested exactly at that point, the third is $8\varepsilon$ with the single state $\Phi_{22}^+\chi_{00}$.
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