= Solution
Use collisional <inverse bremsstrahlung> in a weakly coupled, unmagnetized plasma. The specified density is $n_e=10^{26}\,\mathrm{m^{-3}}$, and $k_BT=1000\,\mathrm{eV}$. The electron <Debye length> is $2.35\times10^{-8}\,\mathrm m$, the <plasma frequency> is $5.64\times10^{14}\,\mathrm{s^{-1}}$, and the laser angular frequency is $\omega=2\pi c/\lambda=1.79\times10^{15}\,\mathrm{s^{-1}}$. Thus $X=\omega_p^2/\omega^2\simeq0.099<1$: the plasma is underdense and a propagating wave exists.
Estimate $b_{\min}$ as the larger of the classical Coulomb-deflection length $Ze^2/(4\pi\epsilon_0k_BT)$ and the electron quantum length $\hbar/\sqrt{m_ek_BT}$. Both are about $9\times10^{-12}\,\mathrm m$ here, giving $\ln\Lambda_C\simeq8$. With the usual thermal collision convention,
$$
\nu_{ei}\simeq2.91\times10^{-6}\frac{n_e[\mathrm{cm^{-3}}]Z\ln\Lambda_C}{T_e[\mathrm{eV}]^{3/2}}\,\mathrm{s^{-1}}\simeq4.4\times10^{11}\,\mathrm{s^{-1}}.
$$
The $Z$ factor follows from $n_iZ^2=n_eZ$.<NRL Plasma Formulary (2019)>[Coulomb collision-rate normalization and thermal parameters; densities in cubic centimetres and temperatures in eV]{ref}
The weakly collisional <Drude model> gives $\epsilon_r=1-\omega_p^2/[\omega(\omega+i\nu_{ei})]$ for fields proportional to $e^{-i\omega t}$. Expanding at $\nu_{ei}\ll\omega$ yields a real refractive index $n_r=\sqrt{1-X}\simeq0.949$ and intensity attenuation coefficient
$$
\alpha_I=2\operatorname{Im}k\simeq\frac{\nu_{ei}}{c}\frac{X}{\sqrt{1-X}}.
$$
Consequently the intensity e-folding absorption length is
$$
\boxed{L_I=\frac{c\sqrt{1-X}}{\nu_{ei}X}\simeq6.6\times10^{-3}\,\mathrm m,}
$$
i.e. of order a centimetre. Taking a Coulomb logarithm between about 5 and 10 gives roughly $5$–$10\,\mathrm{mm}$. The electric-field amplitude decay length is twice this intensity length. This estimate assumes a uniform slab thick enough to define an absorption length and neglects gradients, nonlinear quiver heating and other absorption mechanisms.
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