 Solution

Write $\Theta =k_{B}T$ in $\mathrm{keV}$. The photon count ratio in d eliminates radius, electron number density and the common normalization:The extra ${\epsilon }_2/{\epsilon }_1$ is essential for a photon number spectrum. HenceThe measured central Compton y parameter and the supplied conversion coefficient then determine a column combination:To determine the two remaining quantities separately, define $\hat{R}=R/(1\mathrm{Mpc})$ and $\hat{n}=n_e/(1\mathrm{c}{\mathrm{m}}^{-3})$. Use $1\mathrm{Mpc}=3.08567758\times 10^{22}\mathrm{m}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $j_0=4.885\times 10^{-51}\mathrm{W}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}\mathrm{s}{\mathrm{r}}^{-1}$, $T_0=1.977\times 10^9\mathrm{K}$, and $x_H=0.76$ in d. With the spectral unit conversion explained there, the $0.2\mathrm{keV}$ count luminosity isExplicitly the numerical coefficient isSince $\hat{R}\hat{n}=0.00200087$, substitution gives ${\hat{R}}^3{\hat{n}}^2=\hat{R}(0.00200087{)}^2$, soRounding consistently with the supplied coefficient $0.004$, the cluster has $T\simeq 10^7\mathrm{K}$, $R\simeq 1\mathrm{Mpc}$ and $n_e\simeq 2\times 10^{-3}\mathrm{c}{\mathrm{m}}^{-3}$. Substituting the unrounded values into both X-ray count luminosities gives $2.8800\times 10^{54}$ and $4.4000\times 10^{50}\mathrm{ke}{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$, and the central Compton y parameter returns $1.3800\times 10^{-5}$. If the observed $y$ were a projected-disk average, its central value would be $3/2$ times larger and this same inversion would give $R$ smaller by $4/9$ and $n_e$ larger by $27/8$; specifying the sightline is therefore part of the inference.