A 20 ~ H inductor coil is connected to a 10Ω & resistance in series...

Rate of dissipation of energy in resistor $= {i}^{2}R$
Rate of energy stored in inductor $= \dfrac{d}{dt}\left(\dfrac{1}{2}L{i}^{2}\right) = Li\dfrac{di}{dt}$
${i}^{2}R = Li\dfrac{di}{dt}$
$\dfrac{di}{dt} = i\dfrac{R}{L} \ldots \left(i\right)$
In $L - R$ circuit:

$i = {i}_{0}\left(1 - {e}^{- \dfrac{t}{τ}}\right) ( ∴ τ = \dfrac{L}{R} = 2 )$
$\dfrac{di}{dt} = \dfrac{{i}_{0}}{τ}{e}^{- t / τ}$
From equation (i),

$\dfrac{{i}_{0}}{τ}{e}^{- t / τ} = {i}_{0}\left(1 - {e}^{- \dfrac{t}{τ}}\right)\dfrac{R}{L}$
${e}^{- t / τ} = 1 - {e}^{- t / τ}$
${e}^{- \dfrac{t}{τ}} = \dfrac{1}{2} ⇒ t = τ\ln 2$
$= 2\ln 2$