A coil of self inductance 10mH and resistance of 0.1Ω is connected...

The instantaneous current in an LR circuit is given by the equation :
(instantaneous current) $I = {I}_{s}\left(1 - {e}^{- \dfrac{Rt}{L}}\right)$

Where $I, {I}_{s}, R, t, L$ are the instantaneous current, saturation current, resistance, time and inductance respectively.

Given that the current becomes so percent of saturation current.
$⇒ \left(0.8\right){I}_{s} = {I}_{s}\left(1 - {e}^{- \dfrac{Rt}{L}}\right)$

$R = 0 . 1\text{Ω} + 0 . 9\text{Ω} = 1\text{Ω}$
$⇒ 0.8 = \left(1 - {e}^{- \dfrac{1 \times t}{10 \times {10}^{- 3}}}\right)$
$⇒ 0.8 = 1 - {e}^{- 100t}$
Therefore, the time is:
$t = \dfrac{\ln \left(\dfrac{1}{1 - 0.8}\right)}{100} = \dfrac{\ln \left(5\right)}{100} = 0.016s$