= Solution
For an independent count with a <Poisson distribution>, collected for exposure $T_i$ at rate $R_i$, the expected count is $N_i=R_iT_i$ and its <standard deviation> is $\sqrt{N_i}$. The fractional rate uncertainty is consequently
$$
\frac{\sigma_{R_i}}{R_i}\simeq\frac1{\sqrt{N_i}}.
$$
A target of $0.15$ requires $N_i\simeq(0.15)^{-2}=44.44$ counts at each proper-time setting. The <exposure time for a Poisson rate at fixed fractional precision> is therefore $T_i=1/(0.15^2R_i)$, giving
$$
\begin{array}{r|r}
t\ (\mathrm{ps})&T_i\ (\mathrm{days})\\ \hline
0.25&0.03319\\
0.75&0.12346\\
1.25&0.40404\\
1.75&1.43369\\
2.25&3.17460\\
2.75&14.81481
\end{array}
$$
Since the settings are acquired independently rather than simultaneously,
$$
\boxed{T_{\mathrm{total}}=\frac1{0.15^2}\sum_i\frac1{R_i}
\simeq19.98\,\mathrm{days}\simeq20\,\mathrm{days}.}
$$
Nearly three quarters of the exposure is spent at the latest, lowest-rate setting. Proper time in picoseconds labels the decay distribution; it is not the laboratory acquisition time in days. This estimate assumes counting statistics dominate, with no additional background-subtraction uncertainty or acquisition overhead.
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