= Solution
Model the room's decaying <acoustic waves> as a weakly damped <resonator>. The <quality factor from ringdown> is related to stored <energy> $E$ by
$$
Q=\omega\frac{E}{-dE/dt},\qquad E(t)=E(0)e^{-\omega t/Q},\qquad A(t)=A(0)e^{-\omega t/(2Q)},
$$
where $A$ is the pressure <wave amplitude> and $\omega=2\pi f$. The measured <frequency> $f=440\,\mathrm{Hz}$ gives $\omega\simeq2.76\times10^3\,\mathrm{s^{-1}}$. If the stated $5\,\mathrm{s}$ is interpreted as an approximate <wave amplitude> <exponential relaxation time>,
$$
\boxed{Q\simeq\frac{\omega\tau_A}{2}=\pi(440\,\mathrm{Hz})(5\,\mathrm{s})\simeq6.9\times10^3,}
$$
an estimate of order $10^4$. Taking $5\,\mathrm{s}$ as an <energy> <exponential relaxation time> instead gives $Q\simeq1.4\times10^4$.
There is a real convention ambiguity: “heard to die away” gives neither an initial sound level nor an audibility threshold, so it does not determine either <exponential relaxation time>. More generally, if the <energy> falls by a factor $E(0)/E(t_d)$ in $t_d$,
$$
Q=\frac{2\pi f t_d}{\ln[E(0)/E(t_d)]}.
$$
In room <acoustics>, a <reverberation time> $T_{60}$ usually denotes a <decibel> drop of $60\,\mathrm{dB}$, or an <energy> factor $10^6$. If $5\,\mathrm{s}$ means $T_{60}$, the corresponding estimate is instead $\boxed{Q\simeq2\pi(440)(5)/(6\ln10)\simeq1.0\times10^3}$. The simple decay-time estimate and the conventional <reverberation time> estimate should not be silently equated.
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