= Solution
For $N_p$ <protons>, exposure $T$ and efficiency $\epsilon$, the expected count in this channel is $\mu=\epsilon N_p\Gamma_{\mu\pi^0}T$. A partial <decay rate> at the quoted upper bound gives one event per year when
$$
N_p\simeq6.6\times10^{33},\qquad\boxed{M_p=N_pm_p\simeq1.10\times10^7\ \mathrm{kg}\simeq11\ \mathrm{kilotonnes}}.
$$
More generally, $M_p\simeq11\ \mathrm{kilotonnes}\,\mu/[\epsilon(T/\mathrm{yr})]$ at the limiting rate. This is an estimate of the required proton inventory, not necessarily the whole detector <mass>. For water, each molecule contains ten <protons> and has <mass> approximately $18$ atomic mass units; the corresponding ideal inventory is about $19.7\ \mathrm{kilotonnes}$ of water. Finite efficiency and a useful sample larger than one event increase the required exposure.
An upper limit is not a measured nonzero rate. The true rate could be much smaller, including zero, so no finite target guarantees a measurable signal. The estimate describes sensitivity to a rate near the existing limit.
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