= Solution
Keep the once-per-bond Hamiltonian convention of part (a). Let each sublattice <magnetization> be its <magnetic moment> \i[per total physical volume], so the density of B-sublattice <spins> is $1/(2a^3)$. Writing their moments as $\boldsymbol\mu=g\mu_B\mathbf S$ fixes
$$
\langle\mathbf S_B\rangle=\frac{2a^3}{g\mu_B}\mathbf M_B.
$$
The exchange <energy> of one A <spin> is $6J_1\mathbf S_A\cdot\langle\mathbf S_B\rangle=-\boldsymbol\mu_A\cdot\mathbf B_{\rm mf}$, giving
$$
\boxed{\mathbf B_{\rm mf}=-\lambda\mathbf M_B,\qquad
\lambda=\frac{12J_1a^3}{(g\mu_B)^2}>0.}
$$
A convention using $2J_1$ per bond would double this coefficient; defining <magnetization> per sublattice volume instead changes its normalization. The definitions above make the total <magnetization> $\mathbf M=\mathbf M_A+\mathbf M_B$.
For equal fixed sublattice magnitudes $M_0$, put $\mathbf M_A=M_0(\sin\theta,0,\cos\theta)$ and $\mathbf M_B=M_0(\sin\theta,0,-\cos\theta)$, with the external field along $+x$. The effective A-site field is $(B_\perp-\lambda M_0\sin\theta,0,\lambda M_0\cos\theta)$. Its alignment with $\mathbf M_A$ requires $B_\perp=2\lambda M_0\sin\theta$, so
$$
\boxed{\theta=\arcsin\frac{B_\perp}{2\lambda M_0}\simeq\frac{B_\perp}{2\lambda M_0},\quad
M_\perp=2M_0\sin\theta=\frac{B_\perp}\lambda,\quad
\chi_\perp=\frac{M_\perp}{H_\perp}=\frac{\mu_0}\lambda.}
$$
The last equality uses the externally imposed vacuum field $B_\perp=\mu_0H_\perp$ and neglects demagnetizing corrections. Canting costs exchange <energy> but need not change the magnitude of either sublattice moment. The result holds before saturation and assumes the ordered axis is maintained when defining a transverse response.
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