* World electricity production: 25000TWh/year https://en.wikipedia.org/wiki/Electricity_generation
* Energy per Helium atom: 17.6MeV https://en.wikipedia.org/wiki/Deuterium%E2%80%93tritium_fusi...
So per year (with a 100% efficiency [1]) you would produce as a side effect 25000TWh/17.6MeV ~= 3E31 Helium atoms.
The Avogadro constant is 6,022E23 and that produces 22.4 liters of gas, so you get 1.2E9 liters, that is 1.2E6 cubic meters. Let's round that to 1 million cubic meters per year.
The worldwide production of Helium is 140 million cubic meter per year https://www.statista.com/statistics/925214/helium-production...
[1] A 100% efficiency is too optimistic. The second law and the nasty details of the real world reduce the efficiency a lot. Let me guess 50% from https://en.wikipedia.org/wiki/Energy_conversion_efficiency#E... But this is good! If you waste half of the produced energy, you need the double of fusion plants, and that means the double of Helium balloons.
If all power generated right now was from D+T fusion, it would generate about 8.2% of the current helium consumption. We consume 153,596 TWh of thermal energy per year [1]. Each D+T reaction releases 17.59 MeV [2]. Multiply by the atomic mass of He4 and divide by Avogadro's number to get the mass of He4 produced per energy produced. Divide by the density of He4 at STP to get volume of He4 produced per second [3]. Divide by the consumption of He4 to get the ratio of He4 produced to He4 consumed [4].
(153596 TWh / year) / (17.59 MeV) / (avogadro's number 1/mol) * (4.002602 g/mol) / (0.1786 g / liter) / (88 million m^3 / year)
https://www.wolframalpha.com/input/?i=(153596+TWh+%2F+year)+...
1. https://ourworldindata.org/energy-production-and-changing-en...
2. http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/fusion.htm...
3. https://www.engineeringtoolbox.com/gas-density-d_158.html
4. https://link.springer.com/article/10.1007/s11053-017-9359-y
For the foreseeable, we use HTS materials, that can superconduct at LN2 temperatures, with helium.