= Solution
Use normalized plane-wave <spin> orbitals $V^{-1/2}e^{i\mathbf k\cdot\mathbf r}\chi_\sigma$ and fill a separate <Fermi sphere> for each <spin> species. Each <spin> state counts once, so
$$
n_\sigma=\int_{|\mathbf k|<k_{F\sigma}}\frac{d^3k}{(2\pi)^3}
=\frac{k_{F\sigma}^3}{6\pi^2}.
$$
Its <kinetic energy> per <volume> is
$$
\frac{T_\sigma}{V}=\int_{|\mathbf k|<k_{F\sigma}}\frac{d^3k}{(2\pi)^3}
\frac{\hbar^2k^2}{2m_e}
=\frac{\hbar^2 k_{F\sigma}^5}{20\pi^2m_e}
=A n_\sigma^{5/3},
\qquad
\boxed{A=\frac{3\hbar^2}{10m_e}(6\pi^2)^{2/3}.}
$$
The spin-independent contact interaction has uncancelled direct <energy> $E_H/V=\gamma n^2/2$, with $n=n_\uparrow+n_\downarrow$. By the imposed background assumption this direct contribution cancels. For the Fock term the contact delta function sets the two spatial positions equal; its <spin> factors then select equal spins. The local one-body <density> matrix for each <spin> has diagonal value $n_\sigma$, giving
$$
\frac{E_X}{V}=-\frac\gamma2\sum_\sigma n_\sigma^2.
$$
Combining these results gives the <contact-interaction Hartree-Fock spin energy>
$$
\boxed{\frac{E_0}{V}=A(n_\uparrow^{5/3}+n_\downarrow^{5/3})
-\frac\gamma2(n_\uparrow^2+n_\downarrow^2).}
$$
Here $\gamma$ has units of <energy> times <volume>. The result is the model with the stated background subtraction; an isolated contact-interacting gas without that subtraction retains its direct contribution.
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