I'm with you all the way, apart from 101.19/115 equaling 10.05. I feel awful for pointing that out.Easy method, use the adiabatic equation to find out the size
S = ✓ I^2 * t / k
I = the current required to operate the device
t = disconnection time required for the device
k = factor given for the cable selected - Typically 115
So for argument's sake, let's say we have a fault current of 320 (taken from your trip curves) a time of 0.1 (taken from your device operating times) and a K factor of 115 (taken from the tables following the trip times)
S = ✓ I^2 * t / k
S = ✓ 320^2 * 0.1 / 115
S = ✓ 102,400 * 0.1 / 115
S = ✓ 10,240 / 115
S = 101.19 / 115
S = 10.05
As there is no conductor with a CSA of 10.5, you select the next size up, in this case 16mm^2
The above is purely and illustration and values may not be accurate but it serves to show how to lay out the equation so that you can understand what is going on at each step.
Hope this helps
Sorry