Let us, per the article's title, do the math...
* Total energy consumption = 10^26 watts (hope the caret comes through HN formatting)
* Assume some magical generation and transport mechanisms which don't involve energy gradiants (avoiding pesky thermodynamic realities), and are 99.9% efficient.
* Total waste energy is 10^23 watts.
* Radiators are on Earth, and we dedicate 1/2 of the entire surface area (not just land) to radiators. Earth has about 5.1 x 10^8 km^2, so say 2.5 x 10^14 sq. meters.
* Radiators for the waste heat would be dumping 10^23 W / 2.5 x 10^14 m^2 = 4 x 10^8 watts per square meter. That's 400 megawatts. Per square meter.
* That's a black-body temperature of 9,165 K (the Sun's surface is 5,772 K). Peak wavelength is 316 nm. So half the Earth's surface is a glowing UV light pointed at the sky. I don't think we'd have to worry about the ozone layer disappearing. (A very bright, inefficient UV lamp--plenty of infrared, hard UV, and soft X-rays to go around!)
* OK, so put the power plants in geostationary orbit. Say 1 million power stations, each with radiators 1 square kilometer.
* 1e6 stations * 1e9 square meters = 1e15 sq meters.
* So, that's only 1e8 watts/meter! Hmm. Still 6,480 K, but peak is blue-violet rather than UV. Still lots of UV and infrared, but it's space, who cares. Just don't cross the beams (ha ha, meaning no deep-space traffic). And mind the Moon! (Wouldn't want to give anbody on earth a reflected sunburn).
* The radiation pressure from the waste engery for one power stations would be about 730 kN (160,000 pounds). Would need counter-balanced beams.
* Geostationary orbit is about 26,000 miles. The surface area of a sphere that big is 2e16 square meters. So about 1/20th of the sky would be covered with powerplant radiators.
And this is just for the hypothetical 0.1% waste energy. Multiply all of the above by 1,000 for the actual used energy. So you'd have half the Earth's surface as power receivers slurping up 400 gigawatts/square meter of X-rays and hard UV. Each power station would see a force of 739 mega-newtons; you'd have to have a counter-balancing beam pointing away from Earth. And then you'd need to radiate away all that energy after it was used.
Edit: Oh, and while 1/2 of the Earth is receiving 400 GW/m^2, the other half is using that amount. Which would be the real limit. I don't think anybody would carry around a 10 GW iPhone 237+.