= Solution
Put $A=2\mu aF_0k/(cf)$. In the specified near-wall limit the outer <temperature> is $T-T_0\simeq A/(bz)$, so
$$
q_C\simeq\frac{2\kappa A}{bz^3},\qquad
\frac{|q_C|}{q_R}\simeq\frac{4\kappa\mu a}{\rho cfbz^3}
=\frac{4\kappa\nu a}{cfbz^3},\qquad
\chi=\frac{\kappa}{\rho c}.
$$
Thus the <near-wall radiative heating layer> has scale
$$
\boxed{0<z\lesssim\delta_C,\qquad
\delta_C=\left(\frac{4\kappa\mu a}{\rho cfb}\right)^{1/3}
=\left(\frac{4\kappa\nu a}{cfb}\right)^{1/3}.}
$$
The comparison is an estimate of the breakdown of the outer solution, not a new inner <temperature> solution. It requires $\delta_C\ll b$ and $k\rho\delta_C\ll1$; neglecting longitudinal relative to transverse <thermal conduction> additionally requires $\delta_C\ll a$ in the region to which the estimate is applied.
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