= Solution
Neglect <electron mass> and nuclear binding corrections when converting <mass density> $\rho$ into <hydrogen> and <helium> <number densities>. With $x=x_H$,
$$
n_H=\frac{x\rho}{m_p},\qquad n_{\rm He}=\frac{(1-x)\rho}{4m_p},\qquad
n_e=n_H+2n_{\rm He}=\frac{(1+x)\rho}{2m_p}.
$$
The <thermal free-free emissivity> adds over ion species with weights $Z_i^2$. Consequently
$$
\sum_i Z_i^2n_i=n_H+4n_{\rm He}=\frac\rho{m_p}=\frac{2n_e}{1+x}.
$$
With each <Gaunt factor> set to one, the mixture has
$$
\boxed{j_\nu=j_0\frac{2}{1+x_H}
\left(\frac{n_e}{10^{-3}\,\mathrm{cm}^{-3}}\right)^2
\sqrt{\frac{T_0}{T}}e^{-h\nu/(k_BT)}.}
$$
At fixed $n_e$, pure <hydrogen> gives the composition factor $1$, and pure <helium> gives $2$. The latter check follows independently from $n_{\rm He}=n_e/2$ and $Z^2=4$; simply replacing $Z$ by an average charge would lose this weighting.
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