= Solution
In the <small-angle approximation> the <physical pendulum> is a <harmonic oscillator> with $\omega=2\pi/T$. Its classical <energy> is $E=m\omega^2A^2/2=3.55\times10^{-3}\,\mathrm J$. Equating this to $(n+1/2)\hbar\omega$ gives
$$
\boxed{n\simeq\frac{m\omega A^2}{2\hbar}=5.36\times10^{30}}.
$$
An oscillating bob is not a single stationary <energy eigenstate>. An ideal <coherent state> has a <Poisson distribution> of oscillator occupations centred on this <mean>, with relative spread $1/\sqrt{\bar n}$, exceptionally small here. This <quantum occupation scale of a classical oscillator> fixes the excitation scale without requiring the real, damped clock to be a pure <coherent state>.
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