a) Solve , with I(0) = 0 (which is the meaning of "initially there is no current in the circuit"), and L, R, and E0 constant. If you don't know how to do that, then your trouble is that you don't understand the basics of differential equations. I'll assume that you do know how to do that, so that you've solved for I(t).

b) I believe that the "steady state solution" is asking what I(t) is "asymtopically equivalent to" for large time values. If you did (a) and understand what this means, it's not hard to determine.

Ex: If I(t) were a polynomial in t, then the steady state solution would be the highest power term of the polynomial.

Ex: If I(t) = 3t^2 -4t +7, then I(t) ~ 3t^2, because -4t +7 are eventually a miniscule fraction of I(t) once t is large.

Ex: If I(t) = 2t + 5sin(t), then I(t) ~ 2t, since 5sin(t) is a minscule fraction of I(t) when t is large (5sin(t) bounces between -5 and 5).

Ex: If I(t) = (3t +2)/(t+1), then I(t) ~ 3. That's because I(t) = (3t + 2)/(t+1) = (3(t+1) -3 +2)/(t+1) = (3(t+1)-1)/(t+1) = 3 - (1/(t+1)), and 1/(t+1) goes to 0 as t goes to infinity.

Ex: Any term going to 0 is discarded from the steady state solution. If I(t) = f(t) + g(t), and limit {t -> infinity) g(t) = 0, then the steady state solution for I(t) is the same as the steady state solution if I(t) has equalled *just* f(t). For the steady state, you can discard terms going to 0.

c) In part (a), you solved for a function I(t). In part (b), you determined a new function of t, the steady state function of I(t), let's call it SS(t).

(In other words, I(t) ~ SS(t)). Now part (c) asks "When will the current be (1-e^-1) times the steady state current?". Suppose that specific time(s) is labelled t'.

Then (c) is asking: "Find t' such that I(t') = (1-e^-1) SS(t')." So simply write down that equation and solve it for t'.

Note that, in general, there could be more than one solution. Note also that the domain here is t >= 0, so negative solutions must be discarded.