In the given circuit diagram (figure), key K is switched on at t = 0. The ratio of current i through the cell at t = 0 to that at $t=\infty$ will be

At t = 0, the branch containing L will offer infinite resistance while the branch containing the capacitor will be effectively a short circuit. 

Hence,${\left(i\right)}_{t=0}=\frac{\epsilon }{R}$. Similarly at $t=\infty$, L will offer zero resistance whereas C will be an.open circuit. 

Hence effective resistance $=R+\frac{6+3R}{6+3}=3R{\left(i\right)}_{t=\infty }=\frac{\epsilon }{3R}$

The required ratio $=\frac{\epsilon }{R}\times \frac{3R}{\epsilon }=3:1$