1. How many minutes could a consumer with $3~{\rm W}$ get supplied by an ideal battery with $10~{\rm kWh}$ ?
Solution steps
\begin{align*}
W &= 10~{\rm kWh} &&= 10'000~{\rm Wh}\\ \\
t &= {{W }\over{P }}
&&= {{10'000~{\rm Wh}}\over{3~{\rm W}}} \\
&&&= 3'333~{\rm h}= 200'000~{\rm min} \quad (= 139~{\rm days})
\end{align*}
Result
\begin{align*}
t = 200'000~{\rm min}
\end{align*}
2. Hard: Why is a real Lithium-ion battery with $10~{\rm kWh}$ unable to provide the given power for the calculated time?
Result
There are additional losses:
The battery has an internal resistance. Depending on the current the battery provides, this leads to internal losses.
The internal resistance of the battery depends on the state of charge (SoC) of the battery.
The wires also add additional losses to the system.