There's more details and subtleties to available fault current calculations than I'm familiar with, but here's a basic answer to start with:
Say you know that at a certain point in the electrical system, if you short two wires with a nominal 240V potential between them, you'll get a fault current of 21,188A. [APS's response implies this is true at the terminals of the PV breaker in the service panel, and I question whether that's a sharp number or just a conservative estimate; if the real value is lower, that would make things easier for you.]
Then Ohm's law tells you that the source impedance at that point is just R = V/I = 240V / 21,188A = 11.3 milliohms. To get the available fault current down to 10,000A, you'd need a source impedance of R = V/I = 240V / 10,000A = 24 milliohms. So to do that you need to add 12.7 milliohms of impedance.
If you want to do that just relying on the #8 Cu conductors between the service panel and the unfused disconnect, then per Chapter 9 Table 9 in the NEC #8 Cu has an impedance of 0.69 milliohms/ft (for 3 wires in PVC conduit at a 0.85 PF; whether those conditions are reasonable to use are one of the subtleties I'm unclear on ). That means to add 13 milliohms, you'd need a roundtrip conductor length of 12.7/0.69 = 18.4 ft.
So this simple calculation says that you just need to move your PV disconnect to be 9.2 ft or more of one-way conductor run from the service panel, rather than only 6ft. However, I'm unclear on whether considering a 120V line to ground would be the controlling case with a different answer, or if there's something else I'm overlooking. So hopefully someone more experienced will comment.
Cheers, Wayne