= Solution
At distance $r$, one grain intercepts power through its projected area and emits through its full surface:
$$
P_{\rm abs}=\pi a^2\frac{L}{4\pi r^2},\qquad P_{\rm em}=4\pi a^2\sigma_{\rm SB}T_d^4.
$$
The <blackbody dust sublimation radius> follows by setting $T_d=T_{\rm sub}$:
$$
\boxed{r_{\rm sub}=\left(\frac{L}{16\pi\sigma_{\rm SB}T_{\rm sub}^4}\right)^{1/2}\simeq1.62\times10^{15}\,\mathrm m\simeq0.0526\,\mathrm{pc}.}
$$
Here the supplied luminosity is $3.8\times10^{37}\,\mathrm W$. The grain <radius> cancels because both cross-sections scale as $a^2$. The <temperature> exterior to this boundary scales as $T_d\propto r^{-1/2}$. This estimate assumes isotropic illumination, no shielding, and unit absorption/emission efficiency as specified; real grain emissivities can change it. The numerical evaluation uses the external conversion values $\sigma_{\rm SB}\simeq5.6704\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ and $1\,\mathrm{pc}\simeq3.0857\times10^{16}\,\mathrm m$.<CODATA recommended values of the fundamental physical constants 2014 by National Institute of Standards and Technology>[Stefan-Boltzmann constant, rounded at the precision relevant here]{ref}
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