Determine the full load current on the secondary, divide that number by the percent impedance of the transformer, and that gives you a conservative value at the secondary stabs of the xfmr. Depending on the length of the secondary conductors, that will further reduce the available fault current.
For example, 150000/480/1.73/.02=9030A of fault current at the secondary stabs.
You would be safe choosing the lowest AIC of the group.
