= Solution
The infrared and optical transitions share the same upper level 3. Using the supplied infrared photon <energy> gives
$$
n_3=\frac{1.5\times10^{-21}}{(3.3\times10^{-7})hc/(314\times10^{-6})}
=7.185\times10^6\,\mathrm{m^{-3}}.
$$
With the same $T_e,n_e$ and three-level fraction $f_3$ used in the abundance calculation of <4/b>,
$$
\boxed{n(\mathrm S^+)=5.31\times10^7\,\mathrm{m^{-3}},\qquad
\frac{n(\mathrm S^+)}{n_e}=5.31\times10^{-5}.}
$$
This exceeds the optical abundance by a factor $5.53$. Because the upper level is common, its population cancels in the intrinsic emissivity ratio:
$$
\frac{j_{32}}{j_{31}}=\frac{A_{32}\nu_{32}}{A_{31}\nu_{31}}
=2.715\times10^{-6}.
$$
The observed ratio is $1.5\times10^{-5}$. Thus changing only the <electron temperature> or a common level-population correction cannot reconcile these two lines from the same gas. Dust extinction suppressing the optical line is a natural explanation. If the far-infrared attenuation is negligible, the required optical attenuation is
$$
\boxed{A_{671.6}=2.5\log_{10}(5.53)\simeq1.86\ \mathrm{mag}.}
$$
Different apertures, filling factors, optical-depth assumptions or emissivity calibrations could also cause a discrepancy. A total sulfur abundance would still need an <ionization> correction beyond this ionic comparison.
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