Answer
When $~~t = 6.91~\tau_L~~$ the current will be 0.100% less than its equilibrium value.
Work Step by Step
Note that the equilibrium value of the current is $\frac{\mathscr{E}}{R}$
Let $~~t = n~\tau_L$
We can find $n$:
$i= \frac{\mathscr{E}}{R}~(1-e^{-n~\tau_L/\tau_L})$
$0.999 =(1-e^{-n})$
$e^{-n} = 0.001$
$e^n = 1000$
$n = ln(1000)$
$n = 6.91$
When $~~t = 6.91~\tau_L~~$ the current will be 0.100% less than its equilibrium value.
