= Solution
Assume a single-temperature optically thin <local thermodynamic equilibrium> population, negligible line absorption/stimulated corrections, and that the measured line <luminosity> is from the remnant's CO. Each spontaneous $J=1\to0$ decay emits $h\nu_{10}$, giving
$$
N_1=\frac{L_{10}}{A_{10}h\nu_{10}}
=\frac{1.33\times10^{24}}{(7.2\times10^{-8})(6.6261\times10^{-34})(115.3\times10^9)}
=2.42\times10^{53}.
$$
For a <rigid rotor>, $E_J/k_B=T_rJ(J+1)$ with $T_r=h\nu_{10}/(2k_B)=2.767$ K. The degeneracy is $g_J=2J+1$, and the <rotational partition function> is
$$
Z(T)=\sum_{J=0}^\infty(2J+1)e^{-T_rJ(J+1)/T},\qquad
f_1(T)=\frac{3e^{-2T_r/T}}{Z(T)}.
$$
At 20 K the direct sum gives $Z\simeq7.57$, $f_1=0.3005$; the approximation $Z\simeq T/T_r+1/3$ is already adequate. Thus, with molecular <mass> $m_{\rm CO}\simeq28m_u$,
$$
\boxed{M_{\rm CO}=\frac{N_1}{f_1}m_{\rm CO}=3.74\times10^{28}\ \mathrm{kg}\simeq1.88\times10^{-2}M_\odot}.
$$
The remnant age does not enter this optically thin population count. It becomes relevant when estimating emitting volume, density, expansion and optical depth to test the assumptions.
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