Answer
a) $8 .6 4\times 10^{13}\;\rm J$
b) $4.19\times 10^{15}\;\rm J$
c) No.
Work Step by Step
a) We know that the output work (energy) is given by
$$P_{out}=\dfrac{W_{out}}{\Delta t}$$
where $P$ is the power and $t$ is the time.
Hence,
$$W_{out}=P_{out}\Delta t$$
So the output energy in one day is
$$W_{out}=(1000\times 10^6)(24\times 60\times 60)$$
$$W_{out}=\color{red}{\bf 8 .6 4\times 10^{13}}\;\rm J$$
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b) The heat energy extracted from $1\;\rm km^3$ water is given by
$$Q=mc\Delta T$$
where $m=\rho V$
$$Q=\rho V c\Delta T\tag 1$$
Plugging the known;
$$Q= (1000)(1\times 10^9)(4190)(1)$$
$$W_{out}=\color{red}{\bf 4.19\times 10^{15}}\;\rm J$$
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c) According to the U.S. Energy Information Administration, the total primary energy consumption in the United States in 2020 was
$$Q_{tot} \approx \bf 9.80\times10^{20}\;\rm J $$
According to the United States Geological Survey (USGS), the total volume of water on Earth is estimated to be around
$$V=1.386 \times 10^9\;\rm km^3$$
So to find how many cubic kilometers needed to supply USA with energy in one year, we need to solve (1) for $V$.
$$V=\dfrac{Q}{\rho c\Delta T}$$
Plugging the known;
$$V=\dfrac{ 9.80\times10^{20}}{(1000)(4190) (1)}=\bf 2.34\times 10^5\;\rm km^3$$
So the volume of the water needed for this engine is reliable and not huge relative to the volume of the water on Earth.
But still we need to find a hot big resrvior (which could be the hot water on the surface of oceans) and a cold reservoir (which could be the depth water on the bottom of the ocean).
But the temperature difference would give us an efficiency of less than 0.1.
If we assumed that the surface water temperature is 20$^\circ$C and the depth water is 0$^\circ$C, so the efficiency is then
$$\eta=1-\dfrac{T_C}{T_H}=1-\dfrac{0+273}{20+273}$$
$$\eta=\bf 0.068$$
and this is the maximum efficiency we can get and hence the real effcicency is much less than that.
So it is a bad invest.
