= Solution
For the ideal guaranteed-delay model, the total untreated yearly input at $250$ operating days is
$$
D_{y,0}=250(7.992\times10^{11})=1.998\times10^{14} \mathrm{Bq}.
$$
Since the discharge is a decreasing function of the <residence time>, the minimum is obtained at equality:
$$
D_{y,0}e^{-\lambda\tau_{\min}}=10^{12},\qquad
\tau_{\min}=\frac{\ln(199.8)}{\lambda}=9308.46 \mathrm s
=\boxed{155.14 \mathrm{min}=2.586 \mathrm h}.
$$
This is $7.642$ <half-lives> and corresponds to \b[$3.444$ L of carbon-dioxide-equivalent delay volume]. The three-hour choice provides margin; operating every day instead would require at least $166.22$ minutes under the same assumptions.
This minimum is conditional on a guaranteed <residence time>. More generally the surviving fraction is
$$
\frac{a_{\rm out}}{a_0}=\int_0^\infty E(t)e^{-\lambda t}\,dt,
$$
where $E(t)$ is the <residence time distribution>. Because the <exponential function> is <convex function>, a spread of transit times gives more surviving <radioactive activity> than <plug flow> at the same mean time. A perfectly mixed continuously flowing vessel has a steady-state survival fraction $1/(1+\lambda V/Q)$, not $e^{-\lambda V/Q}$. Thus no unique minimum \i[tank filling time] follows without specifying the filling, retention and emptying schedule. The boxed minimum applies to the explicitly specified delay system, and must not be represented as approval of an unspecified tank.
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