The current is increasing and flowing left to right in the inductor then left end of inductor will have high potential and right end of inductor will be at low potential. Applying Kirchhoff's loop law ε−VL−iR=0⇒ε−Ldtdi−iR=0 6.0−L×103−1.5×2=0 ⇒L=1036.0−3.0=3.0×10−3H,
Hence L=3mH