oltage across the inductor in the circuit is proportional to the rate of current flow through the inductor. For this first calculate the current flowing through the branch containing L and R2. Then differentiate current i with respect to time t.
In the branch containing L and R2, i=E/R2 1−e−R2t/L
Differentiating current, i with respect to time t we get,
⇒ di/dt = ER2e−R2t/L. R2/L = E/Le−R2t/L
Therefore, VL = L di/dt =Ee−R2t/L = 12e−5tV
Then the potential drop across L as a function of time is 12e−5tV.