Initially, when steady state is achieved,

$i=\frac{E}{R}$

Let $E$ is short circuited at a $t=0.$ Then

At $t=0,$ $i_0=\frac{E}{R}$

Let during decay of current at any time the current

flowing is $-L\frac{di}{dt}-iR=0$

$\Rightarrow \frac{di}{i}=-\frac{R}{L}dt$

$\Rightarrow \underset{i_0}{\overset{i}{\int }}\frac{di}{i}=\underset{0}{\overset{t}{\int }}-\frac{R}{L}dt$

$\Rightarrow {\log }_e\frac{i}{i_0}=-\frac{R}{L}t$

$\Rightarrow i=i_0{e}^{-\frac{R}{L}t}$

$\Rightarrow i=\frac{E}{R}{e}^{\frac{R}{L}t}=\frac{100}{100}{e}^{\frac{-100\times {10}^{-3}}{100\times {10}^{-3}}}=\frac{1}{e}$

KEY CONCEPT : The charging of inductance given

by, $i=i_0\left(1-e^{-\frac{Rt}{L}}\right)$

$\frac{i_0}{2}=i_0\left(1-e^{-\frac{Rt}{L}}\right)\Rightarrow e^{-\frac{Rt}{L}}=\frac{1}{2}$

Taking log on both the sides,

$-\frac{Rt}{L}=\log 1-\log 2$

$\Rightarrow t=\frac{L}{R}\log 2=\frac{300\times {10}^{-3}}{2}\times 0.69$

$\Rightarrow t-0.1\sec .$

$e=-\frac{\mathrm{\Delta }\varphi }{\mathrm{\Delta }t}=\frac{-\mathrm{\Delta }\left(LI\right)}{\mathrm{\Delta }t}=-L\frac{\mathrm{\Delta }I}{\mathrm{\Delta }t}$

$\therefore$ $\left|e\right|=L\frac{\mathrm{\Delta }I}{\mathrm{\Delta }t}\Rightarrow 8=L\times \frac{4}{0.05}$

$\Rightarrow L=\frac{8\times 0.05}{4}=0.1H$