 Solution

Room temperature corresponds to about 0.026 eV, or $2.6\times 10^{-5}$ keV. Read the left-hand branch there and draw a horizontal line to the rising hot branch. The original graph gives a return temperature of order $\boxed{k_B{T}^{\prime }\sim 1\text{–}2\mathrm{keV}}$, with graph-reading uncertainty, not the temperature of minimum conductivity.

At full ionization $n_e=13\rho /(27m_u)\simeq 7.8\times 10^{29}$ m${}^{-3}$. Inserting the inferred temperature into $N_D=(4\pi /3)n_e[{ϵ}_0{k}_B{T}^{\prime }/(n_e{e}^2){]}^{3/2}$ gives a Debye-sphere occupancy of order $60$–$170$. This is much greater than one and supports a collective weak-coupling description more than the conductivity-minimum region does. The approximate collision-time formula and omitted Coulomb logarithm do not warrant forcing the entire plotted conductivity curve to have its exact prefactor.