= Solution
Use the explicit convention $H=J_1\sum_{\langle ij\rangle}\mathbf S_i\cdot\mathbf S_j+J_2\sum_{\langle\langle ij\rangle\rangle}\mathbf S_i\cdot\mathbf S_j$, counting every bond once and taking dimensionless <spins> of length $S$. On the <simple cubic lattice> there are six nearest and twelve face-diagonal next-nearest neighbours.
For type I all six nearest bonds are antiparallel and all twelve next-nearest bonds parallel, so $E_I/N=S^2(-3J_1+6J_2)$. Choose the parallel direction of type II as $z$. It has two parallel and four antiparallel nearest bonds, and four parallel $xy$ face diagonals but eight antiparallel $xz,yz$ diagonals. Thus $E_{II}/N=S^2(-J_1-2J_2)$ and
$$
\frac{E_{II}-E_I}{N}=2S^2(J_1-4J_2).
$$
Consequently, for the stated antiferromagnetic exchanges,
$$
\boxed{\text{type I: }J_2/J_1<1/4;\quad
\text{type II: }J_2/J_1>1/4;\quad\text{degenerate at }1/4.}
$$
This comparison is also consistent with the full classical spiral <energy>. With $u_i=\cos(q_ia)$,
$$
E(\mathbf q)/(NS^2)=J_1(u_x+u_y+u_z)+2J_2(u_xu_y+u_yu_z+u_zu_x).
$$
It is affine in each $u_i\in[-1,1]$, so some minimum occurs at a cube vertex, giving the competing collinear states above. At the boundary, the coefficient of $u_z$ vanishes for $u_x=u_y=-1$, producing the larger degeneracy $(\pi/a,\pi/a,q_z)$ and permutations. It is not an isolated two-state crossing. The first panel of the model figure below shows the competing energies.
\Image[/past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2018-magnetic-and-superconducting-response.png]
{title=Competing antiferromagnetic energies, mean-field susceptibilities, a type-II phase diagram and vortex cores}
{description=Original illustrative models. Susceptibilities use spin-one-half mean-field theory, fields in the phase diagram are normalized, and the order-parameter profile is a schematic cut through separated vortex cores.}
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