Answer
No.
Work Step by Step
If steady-state conditions existed, $i_{L}$ would be equal to the steady-state current through $R_{2}$, i.e. no induced e.m.f (electromotive force) in the opposite direction.
Current through $R_{2} = \frac{V_{S}}{R_{2}}=\frac{12}{6*10^{-3}}=2mA$
Since $i_{L}<2mA$, there is a small current flowing in the opposite direction to that from $V_{S}$. By Lenz's Law, this current was generated by an e.m.f. in the same direction, i.e. an induced e.m.f which opposed the direction of $V_{S}$ (Faraday's Law). This suggests that the inductor is active and hence that steady-state conditions did not exist just before the switch was opened.
