 Solution

Put $\Delta =R-r\ll R$. The near-surface mass density and mean free path from a and b areA simple local escape estimate sets the remaining distance equal to the mean free path, $\ell \simeq \Delta$. It givesFor the Sun's parameters this is $\boxed{\Delta R_{\mathrm{esc},\mathrm{local}}\simeq 1.94\mathrm{km},\Delta R_{\mathrm{esc},\mathrm{local}}/R\simeq 2.79\times 10^{-6}}$.

A more precise operational definition uses the outward radial optical depth. In this varying mass density,The chance of an outward radial photon escaping without another scattering is $e^{-\tau }$. Thus the conventional $\tau =1$ last-scattering depth givesThere is no sharp escape surface: $\tau \ll 1$ gives nearly free propagation, and oblique rays have different optical depths. The local $\ell =\Delta$ estimate actually has $\tau \simeq 1/2$ in this model, not $1$; the factor $\sqrt{2}$ is a convention-dependent order-one correction, not a discrepancy in the Thomson cross-section. Neither scattering criterion establishes a thermalization surface.