It's the diameter of the plunger and the stroke on the swash plate that dictates the fuel flow to each injector. Going by the Bosch Diesel Engine Management book, we'll try to decipher the pump codes. Look for the model number on the injection pump, it should look like this:
V E 4 / 9 F 2200 L 12
V = Distributor Injection Pump
E/R = Axial Piston / Radial Piston high pressure pump
4 = Number of high pressure outlets
9 = Pump plunger diameter
F/E/M = Mechanical Governor / Electric Actuator Mechanism / High Pressure Solenoid Valve
2200 = Upper nominal pump speed (half engine speed for 4 stroke engine)
L/R = Counterclockwise / Clockwise rotation (viewed toward pump drive)
12 = Type code
I think the VE pumps on the Cummins are of the 11 or 12 mm variety. In addition, they've got a longer stroke, from memory it was 5.5 mm instead of 4.5. Doing a bit of math:
Area increase of plunger wrt 9mm plunger is = (12/9)^2 = 1.78
Stroke increase of plunger wrt 9mm plunger is = (5.5/4.5) = 1.22
Multiply the two increases and you've got a net fuel flow increase wrt the "smaller" injection pump = 2.17
189 * 2.17 = 410 lb-ft
There's your increase in torque.
As for my original guesstimation of 189 lb-ft, that was based on the basic specification (9 mm plungers - 4.5 mm stroke) VE pump producing 30kW per cylinder at 4000 RPM. That corresponds to more or less 236 lb-ft at the crank. I deducted (rightly or wrongly, you decide) 20% to fall at a wheel torque figure of 189 lb-ft.
p.s. I'm still a bit foggy about the plunger stroke figure. I suppose you'll have a better measured value?