= Solution
For $D=0.001$, the same <duty cycle> model gives $\bar I=25\ \mathrm{\mu A}+0.999(6.1\ \mathrm{\mu A})=31.0939\ \mathrm{\mu A}$. Thus \b[$\bar P=62.1878\ \mathrm{\mu W}$]. The tenfold <reduction> in active fraction reduces average power by only about $8.23$ because the sleep current remains. The ideal limit as $D\to0$ is \b[$VI_s=12.2\ \mathrm{\mu W}$], before leakage elsewhere or wake-up <energy>. Active and sleep contributions become comparable at a <duty cycle> of roughly $I_s/(I_a+I_s)=0.0244\%$; below that scale lowering sleep current is particularly valuable.
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