= Solution
The <duty cycle> $D=t_{\rm active}/(t_{\rm active}+t_{\rm sleep})$ is the fraction of time spent active. At a constant supply voltage and with switching <energy> neglected,
$$
\bar P=V\,[DI_a+(1-D)I_s].
$$
For $D=0.01$, the average current is $0.01(25\ \mathrm{mA})+0.99(6.1\ \mathrm{\mu A})=256.039\ \mathrm{\mu A}$. Thus \b[$\bar P=512.078\ \mathrm{\mu W}\simeq0.512\ \mathrm{mW}$]. Multiplying only the active power by the <duty cycle> would omit the sleep contribution.
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