The current in the resistor has dropped to about 37% of the initial charging current after one time-constant of time has elapsed.
The relevant equation is Ir = (Vsource / R) (e^(-t/RC)) where Vsource is the voltage applied to the RC circuit, R is the resistance in ohms, C is the capacitance in farads, and t is the time elapsed in seconds, e is the base of natural logarithms (about 2.7182818284590452353602874713527), and the caret ^ indicates exponentiation.
Thus, for one time-constant of elapsed time, RC = t and therefore t / RC = 1. The initial current, which is (Vsource / R), is therefore reduced by the factor (e^(-1)) which is about 0.36787944117144232159552377016146 or (roughly) 37%. For more basic information on RC charging (and discharging)
see this link.
Since this is homework, I really shouldn't be telling you any of this. Google is your friend.