Resistor R1 gets the full 12 Volts. It's a parallel circuit. We can calculate the current in Ammeter 2 like this:
I=V/R = 12V/6 Ohms = 2 Amps
Now being in parallel, the diode and R2 combination also get to share 12V.
The key to this is to realise that the diode takes up 0.6 Volts just to get started, which leaves 11.4 Volts for R2. After that, we can just ignore the diode.
So ammeter 3 has this reading: I = V/R = 11.4V/4 Ohms = 2.85 Amps.
Add up the two ammeter readings to get the reading for ammeter 1: 2 + 2.85 = 4.85 Amps
