= Solution
Assume uniform reradiation over the planetary surface, unit thermal <emissivity>, and zero <Bond albedo>. The planet absorbs over its projected disc and emits over its full sphere. <Radiative equilibrium> and the <Stefan–Boltzmann law> give
$$
\pi R_p^2\frac{L_\odot}{4\pi a^2}=4\pi R_p^2\sigma T_p^4,\qquad
T_p=T_\odot\sqrt{\frac{R_\odot}{2a}}.
$$
Using the liquid-water range $273.15\le T_p\le373.15\,\mathrm K$ as the crude thermal criterion gives
$$
\boxed{a_{\rm in}\simeq0.562\,\mathrm{AU},\qquad a_{\rm out}\simeq1.05\,\mathrm{AU}.}
$$
For nonzero <Bond albedo> $A_B$, both radii acquire the factor $\sqrt{1-A_B}$. For example $A_B=0.3$ gives approximately $0.470$ to $0.877\,\mathrm{AU}$. An airless planet need not have uniform surface <temperature>; inefficient heat redistribution makes the dayside hotter and the nightside colder, so a single equilibrium <temperature> is an averaging approximation.
This simple <habitable zone> contains the Earth's orbit under the zero-<albedo> assumption, but also contains Venus's orbit, although its <greenhouse effect> produces an uninhabitably hot surface. Mars lies beyond the calculated outer boundary. These comparisons show the usefulness of the orbital scale and the limitations of a <temperature>-only criterion. The stipulated airless model also omits the <atmospheric pressure> needed for stable surface liquid water: this is an effective thermal window, not a sufficient habitability condition.
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