You make things too complex !
Quick answer without me writing anything down: Il = 4.7V/2.2kohm so roughly 2 mA
More detailed analysis:
The 4.7 V diode is a zener diode, avalanche or however it works doesns't matter.
What matters is that in forward it would behave as a normal diode. But it is operated in reverse, the kathode is at the upper side and the battery supplies 9 V through a resistor to this kathode. This means that there is enough voltage (9V) for the zener to "zener" at 4.7 V. It will now behave like a 4.7 V DC voltage source ! So at the zener's kathode we will have 4.7 V
Yes but R2 I hear you asking. Ha, that's a trick to confuse you ! The other end of R2 is connected to the input of an opamp and what do we know about inputs of opamps ? In general they have a very high input resistance ! Ergo, no current can flow through R2, so basically we can ignore R2, the 4.7 V will still make it to the + input of the opamp.
Now the opamp, I see that it has a negative feedback, the output is fed back to the - input. In such configuration the opamp will try to make the voltage difference between it's inputs 0 (zero). So let's assume that the opamp succeeds in doing so, then there would be 4.7 V also at it's - input and since that is connected to it's output also the output would be at 4.7V. Such a configuration where the output of an opamp is fed straight back to the - input is called a (voltage) buffer. It just copies (buffers) the input voltage at the + input.
So the 4.7 V ends up across R3 therefore
I(R3) = Il = 4.7 V / 2.2k ohm = 2.14 mA