How much uranium would it take to release 7.604548389^30 GeV of energy?

that is 2.6e26 GeV or 4.3e16 joules

wikipedia:

Typical fission events release about two hundred million eV (200 MeV) of energy for each fission event.

so you need 2.6e26 GeV / 200 MeV = 1.33e27

so that number of Uranium atoms is your answer

Change in energy = [Change in mass] [Speed of light]2

or

Delta E = Delta m * c^2 => 7.604548389^30 GeV = Delta m *(3*10^8)^2

=> Delta m = 7.604548389^30 * 10^9 eV / (3*10^8)^2

(1eV=1.6x10^-19 J)

=> Delta m = 7.604548389^30 * 10^9 *1.6x10^-19 J / (3*10^8)^2 = 0.481 kg

1 kilogram = 6.0221415 × 10^26 atomic mass units

0.481 kg = 2.896 * 10^26 amu

Are you assuming a typical fission reaction, or a matter/anti-matter reaction?

If you use pure U-235 and and it completely fissions, it would take 521.32 kg

If you convert it entirely to energy by reacting it with antimatter, it would take 241.1 g (and an equivalent amount of antimatter)

That is a suspiciously exact amount of energy. The requirements to run the flux capacitor, perhaps, or to open your wormhole ?