The usage of helium for the typical amateur balloon refers to ground level conditions, similar to Standard Temperature and Pressure which is 20 Celsius and 1013 millibars pressure.
The 60 million cubic feet of the NASA "Big 60" balloon refers to the fully inflated envelope at its peak (equilibrium) altitude of 159,000 feet. The pressure there is under a millibar, so the gas is expanded over a thousand times in volume over what it was on the surface. The standard atmosphere specifies a mild temperature very near STP, but during the day there must be significant heating to the envelope. I'll just deal with the pressure difference which dominates.
So whittle those 240,000 amateur balloons down to more like 240 as a result of dividing by 1000. And now the low volume, retail customer cost estimate is $600,000 instead of $60 million.
Sanity/error check on the 240,000 balloon answer:
Mark was using as typical a 1200 gram balloon and 250 cubic feet. For helium
that's good for about 8 pounds of payload according to Liftwin.
The NASA instrument payload was 200 kilograms = 441 pounds.
8 x 240,000 = 1,920,000 pounds of payload lift = 960 tons lift! Impossible.
The specific solution is as follows neglecting extra balloon gas heating beyond ambient and giving STP gas volume:
At 159,000 feet, it's .992 millibar and 271 Kelvin. STP is 1013 millibars and 273 Kelvin.
60 * 10**6 * .992 / 1013 * 273 / 271 = .059 * 10**6 = 59,000 cubic feet (for the on-the-ground volume of helium for the Big 60)
Well, if you figure ~250 cu ft for a typical 1200g balloon, around 240,000. That's a lot of ARHAB flights!
If they were paying commercial rates for their helium (which I think is approaching $1/cu ft again), that'd be $60M for their lifting gas for that mission. Somehow I doubt they're paying that much.