= Solution
Write $\Theta=k_BT$ in $\mathrm{keV}$. The <photon> count ratio in <3/d> eliminates <radius>, <electron> <number density> and the common normalization:
$$
\frac{L_{\varepsilon_1}^X}{L_{\varepsilon_2}^X}
=\frac{\varepsilon_2}{\varepsilon_1}
\exp\!\left(\frac{\varepsilon_2-\varepsilon_1}{\Theta}\right).
$$
The extra $\varepsilon_2/\varepsilon_1$ is essential for a <photon> number spectrum. Hence
$$
\boxed{\Theta=\frac{5-0.2}{\ln\!\left[\frac{(2.88\times10^{54})(0.2)}{(4.40\times10^{50})(5)}\right]}
=0.86212\,\mathrm{keV},\qquad
T=1.00045\times10^7\,\mathrm K.}
$$
The measured central <Compton y parameter> and the supplied conversion coefficient then determine a column combination:
$$
Rn_e=\frac{y}{2(0.004)\Theta}
=2.00087\times10^{-3}\,\mathrm{Mpc\,cm^{-3}}\equiv Y.
$$
To determine the two remaining quantities separately, define $\widehat R=R/(1\,\mathrm{Mpc})$ and $\widehat n=n_e/(1\,\mathrm{cm^{-3}})$. Use $1\,\mathrm{Mpc}=3.08567758\times10^{22}\,\mathrm m$, $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$, $j_0=4.885\times10^{-51}\,\mathrm{W\,m^{-3}\,Hz^{-1}\,sr^{-1}}$, $T_0=1.977\times10^9\,\mathrm K$, and $x_H=0.76$ in <3/d>. With the spectral unit conversion explained there, the $0.2\,\mathrm{keV}$ count <luminosity> is
$$
L_{0.2}^X=\left(7.22106\times10^{59}\,\mathrm{keV^{-1}\,s^{-1}}\right)
\widehat R^3\widehat n^2.
$$
Explicitly the numerical coefficient is
$$
\frac{4\pi}{h(0.2)}\frac{4\pi}{3}
(3.08567758\times10^{22})^3j_0\frac{2}{1.76}
10^6\sqrt{\frac{T_0}{T}}e^{-0.2/\Theta}.
$$
Since $\widehat R\widehat n=0.00200087$, substitution gives $\widehat R^3\widehat n^2=\widehat R(0.00200087)^2$, so
$$
\boxed{R=0.9962\,\mathrm{Mpc}\simeq1.00\,\mathrm{Mpc},\qquad
n_e=2.0085\times10^{-3}\,\mathrm{cm^{-3}}\simeq2.00\times10^{-3}\,\mathrm{cm^{-3}}.}
$$
Rounding consistently with the supplied coefficient $0.004$, the cluster has $T\simeq10^7\,\mathrm K$, $R\simeq1\,\mathrm{Mpc}$ and $n_e\simeq2\times10^{-3}\,\mathrm{cm^{-3}}$. Substituting the unrounded values into both <X-ray> count <luminosities> gives $2.8800\times10^{54}$ and $4.4000\times10^{50}\,\mathrm{keV^{-1}\,s^{-1}}$, and the central <Compton y parameter> returns $1.3800\times10^{-5}$. 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.
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