Your PCB will have about 1/2 second thermal time constant between each regulator. [One cm is 1.14 seconds, 2 cm is 4X longer, 3cm is 9X longer)
The heat rise will be approximately uniform, during your 30 second pulse.
IMHO you need some cooling air.
The central region can dump heat in 4 dirctions, (or even 8 what with corners being used) so dead-spot should not matter.
But with 4 degree per watt thermal spreading resistance out from that central region,
and with 30 watts in the central region, you'll have 4 * 30 = 120 degree C rise.
If you redo the PCB, consider 2 ounce/squareFoot for the 2 planes.
By the way, you can model this in SPICE with a 7 by 13 resistor grid (at the 1cm resolution, if you feel that is adequate). Make each resistor be 70 ohms, so you get that 70 degree C per watt thermal response. Then inject 3 amps into each of the interior nodes. Ground the periphery ????? no. That is not valid.
As the other answer suggests, compute the thermal capacity, for each square centimeter and add that to each interior node as a lumped capacitor.
By the way, are your traces wide enough to take the 2,000 amps??? off board?
In the thermal thinking, I use the thermal resistance of standard copper foil (1 ounce / squareFoot) which is 70 degree C per watt per square of foil.
Since there are two inner layers, solid sheets, the lateral Rthermal is 35 degree per watt per square, for any size square.
Now draw a grid, and heat any one of the interior squares. You will see the eight adjacent squares as paths for heat to exit, thus Rthermal drops by 8x, to
4 degree C per watt.
You can include the next set of surrounding squares, eight of them, edges 3x the original size), but the component density suggests the iteration is not needed.