Let's see. If memory serves, that's one item, so that puts 1 unit of metal at, according to that page, 600cm^3. Let's assume a material with similar density to iron, which has a density of 7.874g/cm^3. So, that's approx. 4.7244kg.
Now, judging by the dyson sphere in the game, it appears to be relatively small compared to more traditional ones - a diameter no doubt smaller than Mercury's orbit, and it isn't a solid shell - it's several rings. Let's go with half of Mercury's orbit for the "average" distance, and there appear to be a great number of rings - let's go with 25, and assume they are all the same size. Let's go with a size of, oh, approx. twice that of Earth's diameter for the width of these rings, and a thousand kilometers thick.
Mercury Orbit: 57,909,050km, or 57,909,050,000m
Thickness: 1,000km, or 1,000,000m
Inner edge: 28,954,525,000m
Outer Edge: 28,955,525,000m
Earth's Diameter: 12,742km, or 12,742,000m
Width/"Height": 25,484km, or 25,484,000m
Volume of a cylinder: V=pi(r^2)h
(V(outer)-V(inner))=V(ring)
V(outer)=pi(28,955,525,000m^2)25,484,000m
V(outer)=67,124,390,674,652,783,675,534,678,928.824m^3
V(inner)=pi(28,954,525,000m^2)25,484,000m
V(inner)=67,119,754,375,944,336,110,865,183,436.524m^3
V(ring)=4,636,298,708,447,564,669,495,492.3m^3
Density of iron=7,874kg/m^3
Mass of each ring=36,506,216,030,316,124,207,607,506,370.2kg
25 rings of mass=912,655,400,757,903,105,190,187,659,255kg
1 unit of metal = 4.7244kg
25 rings of mass/1unit = 193,179,112,851,981,861,228,978,845,833.33
Rounded down, 193,179,112,851,981,861,228,978,845,833 metal.
We're going to need more mines, let alone dwarves.
That said, this should totally be in the trader list anyways. Without a price reduction.
Incidentally, could someone give me an estimate as to what the hell you would have to kill to get enough scrap income to build that? It would cost 6,125,669,484,144,528,831,462 metal per second to finish that in one year.
EDIT: I knew I missed something. Added units.