= Solution
Charge neutrality in fully ionized <hydrogen> gives $n_e=n_p$ and $\rho=(m_p+m_e)n_e\simeq m_pn_e$. With only <Thomson scattering>, the <photon> <mean free path> is $\ell=(n_e\sigma_T)^{-1}$. Hence
$$
\boxed{\ell(r)=L\frac{ar}{\sin ar},\qquad
L=\frac{m_p}{\sigma_T\rho_c}=\frac{4m_pR^3}{\pi\sigma_TM},\qquad
\frac LR=\frac{4m_pR^2}{\pi\sigma_TM}.}
$$
Here and below the tiny <electron mass> correction can be retained by replacing $m_p$ by $m_p+m_e$. $L$ is the central <mean free path>, not a <luminosity>. The divergence at $r=R$ reflects the vanishing <electron> <number density>.
Using $M_\odot=1.9885\times10^{30}\,\mathrm{kg}$, $R_\odot=6.957\times10^8\,\mathrm m$, $m_p=1.6726\times10^{-27}\,\mathrm{kg}$, and $\sigma_T=6.6525\times10^{-29}\,\mathrm{m^2}$ gives
$$
\rho_c\simeq4.64\times10^3\,\mathrm{kg\,m^{-3}},\qquad
\boxed{L_\odot\simeq5.42\times10^{-3}\,\mathrm m,\qquad L_\odot/R_\odot\simeq7.79\times10^{-12}.}
$$
These are this profile's values, rather than the <Sun>'s actual central <mass density> or opacity profile.
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