circuit #1 - 600VDC 10,000A Time Constant = 3mS (meaning at 3mS, the first time constant in the charging of an Inductor is 63%, therefore Current = 6300 Amps at 3mS.
circuit #2 - 600VDC 10,000A Time Constant = 10mS (therefore at 10mS, Current = 6300A)
The current will be about 63% of the steady state maximum (the final value), not 65% of the capability of your source. In other words, you have to take the series resistance into account.
circuit #1 - 600VDC 10,000A Time Constant = 3mS :
If your R is 3.3 ohms and L is 10mH, then the time constant is 3ms. Maximum steady state current with 600 volts is 181 A. Assuming you apply a constant (600V) voltage across the LR circuit.
So, after 3 ms the current is roughly 115 A
With circuit #2 the current after 10ms will also depend on the resistance, but assuming 3.3 ohms the current will be the same 115 A. But if you change the the time constant by changing the value of the resistor, then the current after 10ms will be different than in the first circuit after 3ms.
Or, did I understand something wrong and you are using 0.06 ohms resistor in your LR circuit? In that case your calculations are correct.
Is your problem that if you increase the current more slowly, the circuit breaker becomes more unreliable? How do you know that shorter time constant is easier for the breaker? Try to repeat your experiment, but only vary the inductor and keep the resistance constant.