Let Vo1 be the voltage at the output of the first opamp circuit.
Let Vo be the voltage at the output of the second opamp circuit
Considering the opamp to be ideal.
The base currents at the inputs of the opamp is ˜0
The voltages at the inverting input is equal to the voltage at the non-inverting input due to virtual short/ground
Applying KCL at the inverting node of the first opamp circuit
2.4Vo1 + Vo = 2 ----------------- (1)
Applying KCL at the inverting node of the second opamp circuit
21Vo1 - Vo = 0 ------------- (2)
Solving (1) and (2)
23.4Vo1 = 2
Vo1 = 0.085V
Vo = 1.785V
ia = Vo1 / R4 = 0.085 / 3000
ia = 0.0283mA
ia = 28.3μA