At steady state the capacitor will be frilly charged and thus there will be no current in the 10Ω resistance. So the effective circuit becomes
Net current from the 6V battery, I=(2+32×3)+12.86 =1.2+2.86=23=1.5A
Between A and B, voltage is same in both resistances, 2I1=3I2 where I1+I2=I=1.5 ⇒2I1=3(1.5−I1) ⇒I1=0.9A