= Solution
Let the source's <hydrogen-ionizing photon production rate> be
$$
Q_{\rm H}=\int_{\nu_0}^{\infty}\frac{L_\nu}{h\nu}\,d\nu,
$$
where $h\nu_0$ is the hydrogen <ionization energy>. The idealized <Strömgren sphere> has an almost fully ionized interior and a thin <ionization front>. In <photoionization equilibrium>, every ionization is balanced by a <Case B recombination>, so spherical symmetry gives
$$
Q_{\rm H}=4\pi\int_0^{R_S}\alpha_B n_en_p r^2\,dr.
$$
For pure hydrogen of constant <number density> $n$, the interior has $n_e=n_p=n$, and the <Strömgren radius> is therefore
$$
\boxed{R_S=\left(\frac{3Q_{\rm H}}{4\pi\alpha_Bn^2}\right)^{1/3}}.
$$
Now let the effective number of dust grains per hydrogen nucleus be $f_d$, so the dust <absorption coefficient> is $a(r)=f_d\sigma_dn(r)$. If $Q(r)$ is the ionizing-photon rate crossing the sphere of radius $r$, recombinations and dust absorption give the <linear ordinary differential equation>
$$
\frac{dQ}{dr}+a(r)Q=-4\pi r^2\alpha_Bn(r)^2,
\qquad Q(0)=Q_{\rm H},\qquad Q(R_d)=0.
$$
If $f_d$ denotes a dust mass fraction instead, the grain mass and gas mean particle mass are simply absorbed into the effective product $f_d\sigma_d$. Define the <dust optical depth>
$$
\tau_d(r)=\int_0^r f_d\sigma_dn(s)\,ds.
$$
Multiplication by the <integrating factor> $e^{\tau_d(r)}$ and integration to the dusty front gives its governing equation
$$
\boxed{Q_{\rm H}=4\pi\alpha_B\int_0^{R_d}n(r)^2r^2e^{\tau_d(r)}\,dr}.
$$
The <exponential function> weights recombinations at large optical depth by the extra source photons that dust must remove before those photons reach that radius.
For constant $n$, put $a=f_d\sigma_dn$ and $\tau=aR_d$. The elementary <integral>
$$
\int_0^{R_d}r^2e^{ar}\,dr
=\frac{e^\tau(\tau^2-2\tau+2)-2}{a^3}
$$
reduces the equation to
$$
\boxed{Q_{\rm H}=\frac{4\pi\alpha_Bn^2}{a^3}
\left[e^\tau(\tau^2-2\tau+2)-2\right]}.
$$
Equivalently, with $y=R_d/R_S$ and $\tau_S=aR_S$,
$$
\boxed{e^{\tau_Sy}\left[(\tau_Sy)^2-2\tau_Sy+2\right]-2=\frac{\tau_S^3}{3}}.
$$
Its dust-free <limit> is $y\to1$, while absorption makes $R_d\lt R_S$ for nonzero dust abundance.
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