Oh, looks like you did read it right.. excuse me. DE' and DI' are not actually pins but external resistor connection points.
If VCC4 is connected exactly as shown in the block diagram (i.e. it's only connected to the optocoupler anodes), then VCC4 could instead go to 3V3. If as you said, 20mA is the diode current (I didn't see any value), then the diode forward voltage is 1V (20mA * 200ohms). For 3V3 the resistance would become 2V3/0.02A = 115ohms.
If VCC4 cannot be connected in such a manner, then you'd need to use something to drop the voltage (e.g. a few diodes) or a level translator (e.g. a logic IC, or a BJT transistor). Resistance would be the 200 ohms from the datasheet in this case.