= Solution
Identical <spin one-half> particles are <fermions>. By <fermionic exchange symmetry>, the complete spatial-and-spin <wavefunction> must change sign when the two particle labels are exchanged. Define the normalized antisymmetric <spin-one-half singlet state>
$$
\chi_S=\frac{|\uparrow\rangle_1|\downarrow\rangle_2-|\downarrow\rangle_1|\uparrow\rangle_2}{\sqrt2},
$$
and the three symmetric <spin-one-half triplet states>
$$
\chi_+=|\uparrow\rangle_1|\uparrow\rangle_2,\qquad
\chi_0=\frac{|\uparrow\rangle_1|\downarrow\rangle_2+|\downarrow\rangle_1|\uparrow\rangle_2}{\sqrt2},\qquad
\chi_-=|\downarrow\rangle_1|\downarrow\rangle_2.
$$
The <Hamiltonian> is independent of <spin>, so these spin functions carry no additional <energy>.
Both particles can occupy $u_0$ if their <spin> factor is antisymmetric. The unique <ground state> <wavefunction> is
$$
\boxed{\Psi_g(\phi_1,\phi_2)=u_0(\phi_1)u_0(\phi_2)\chi_S
=\frac1{2\pi}\chi_S,\qquad E_g=0,\qquad g_g=1.}
$$
Two equal-spin particles could not share this orbital, by the <Pauli exclusion principle>.
At the first excited <energy>, $E=\epsilon=\hbar^2/(2mR^2)$, one orbital is $u_0$ and the other is $u_s$ with $s=+1$ or $-1$. Define symmetric and antisymmetric normalized spatial functions
$$
\Phi_s^{\pm}(\phi_1,\phi_2)
=\frac{u_0(\phi_1)u_s(\phi_2)\pm u_s(\phi_1)u_0(\phi_2)}{\sqrt2}
=\frac{e^{is\phi_2}\pm e^{is\phi_1}}{2\pi\sqrt2}.
$$
For each $s$, all allowed <wavefunctions> are spanned by
$$
\boxed{\Phi_s^+\chi_S,\qquad
\Phi_s^-\chi_+,\quad\Phi_s^-\chi_0,\quad\Phi_s^-\chi_-,\qquad s=\pm1.}
$$
There are $2(1+3)$ <orthonormal> states, hence \b[first-excited-state degeneracy $g=8$]. In each product, exactly one factor is antisymmetric, ensuring the required total <fermionic exchange symmetry>.
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