The Death Star is 120km across, according to 2000 Meters per Pixel
Using the 4/3 pi r-cubed formula for the volume of a sphere, we find a volume of about 7.2 million cubic kilometers. This is 7,200,000,000,000,000 cubic meters.
Now, we want to get a rough idea of how much steel that implies, so let’s look at the mass per volume of, say, one of the Twin Towers.
According to http://www.physforum.com/What-was-the-weight-of-a-WTC-Tower_4299.html, one of the Twin Towers used about 96,000,000 kg of structural steel.
(By the way, searching for “world trade center facts” sure comes up with a lot of garbage)
According to World Trade Center - Wikipedia, the trade center was 413 meters high. It has 800,000 square meters of floor, so divide that by the number of stories (110) to get a cross-sectional area of about 7200 square meters. Multiplying height by cross-sectional area gives a volume of 3,000,000 cubic meters. Thus, we get a ratio that 3,000,000 cubic meters of a building implies 96,000,000 kg of steel.
Take the volume of the death star and divide it by this ratio to get 230,000,000,000,000,000 kg of steel are in the death star.
So, what’s the current cost of steel per kilogram? According to http://web.ead.anl.gov/uranium/pdf/ducretecosteffec.pdf, it’s about $1.10 per pound.
So, the cost of the steel would be about $230,000,000,000,000,000 ($230 quadrillion).
That’s JUST the cost of the steel, ignoring labor costs. I also ignored the impact of the purchase of that much steel on the cost of steel, and a bunch of other stuff.
For comparison, the world’s GDP is $65.95 trillion (2006 est.) (just google for “world gdp”).
That’s a hell of a sports bar that Darth Vader told the bank he was building.