Returning to our original circuit let's look at the contribution from Voltage source EB.
Because EA has no internal resistance, the EA source is replaced by a short circuit.
As you can see the circuit is symmetrical, so the resistance that the B power supply sees…call it, RTB…the total resistance at EB is also equal to 100 + (100 || 100)… which is 150 Ω therefore, ITB = EB/RTB = 15/150 = 100 mA. (keep in mind that the B power supply polarity is the opposite to the A power supply therefore the current will be flowing out in this direction).
From the current-divider rule, I2B & I3B = 100 mA/2 = 50 mA each.