= Solution
For six hours of operation at the maximum stated carbon-dioxide-equivalent flow, the daily input volume and untreated <radioactive activity> inventory are
$$
V_d=Q(6)(3600)=7.992\times10^{-3} \mathrm{m^3},\qquad
D_{d,0}=a_0V_d=7.992\times10^{11} \mathrm{Bq}.
$$
For the three-hour guaranteed delay, every element of the incoming stream acquires the same <radioactive decay> factor. The <radioactive activity> released per operating day, attributing the delayed tail to its production day, is
$$
D_d=D_{d,0}e^{-\lambda(10800)}=\boxed{1.712\times10^9 \mathrm{Bq}}.
$$
Even on the conservative assumption of operation on all $365$ days,
$$
D_y=365D_d=\boxed{6.248\times10^{11} \mathrm{Bq\,yr^{-1}}},
$$
so the annual discharge is below the supplied $10^{12}$ Bq limit. For $250$ operating days it is $4.279\times10^{11}$ Bq. This sums <radioactive activity> inventories discharged over the year; it does not credit <radioactive decay> in the environment after release. It also retains the final delayed gas after production stops, rather than bypassing the retention system to empty it.
The residence-time qualification matters numerically. A single tank filled for six hours and released immediately, with <radioactive decay> during filling but no later hold, would contain
$$
A_{\rm fill}(21600)=6.502\times10^{10} \mathrm{Bq}.
$$
Repeating that operation on $250$ days releases $1.625\times10^{13}$ Bq, \b[above the limit]. A well-mixed or fill-and-dump system therefore cannot inherit the plug-flow compliance result just because its nominal volume divided by flow equals the selected delay time. The total <nitrogen> flow and any unretained radioactive carbon species must also be included in a real measured release account.
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