Answer
$1.1\times10^4 W$.
Work Step by Step
Find the current used to deliver the power, using P=IV. Then find the power wasted in the resistance.
First, at 12000 volts.
$$I=\frac{P}{V}=\frac{750000W}{12000V}$$
$$P_{wasted}=I^2R=(\frac{750000W}{12000V})^2(3.0\Omega)=11719W$$
Finally, assuming we deliver the power at 50000 volts.
$$I=\frac{P}{V}=\frac{750000W}{50000V}$$
$$P_{wasted}=I^2R=(\frac{750000W}{50000V})^2(3.0\Omega)=675W$$
The substantial savings is the difference, $1.1\times10^4 W$.
