Explanation

(a) In L-R circuit, the growing current at time $t$ is given y $i = i_{0}\left\lbrack 1 - e^{- \frac{t}{\tau}} \right\rbrack$ where $i_{0} = \frac{E}{R}\ \ \ \ and\ \tau = \frac{L}{R}$
$\therefore\ $ Charge passed through the battery in one time constant is
$q = \ \int_{0}^{\tau}{idt = \ \int_{0}^{\tau}{i_{0}\left( 1 - e^{- t/r} \right)dt}}$
$q = \ i_{0}\tau - \left\lbrack \frac{i_{0}e^{- t}}{- 2/\tau} \right\rbrack_{0}^{t}$= $i_{0}\tau + i_{0}\tau\left\lbrack e^{- 1} - 1 \right\rbrack$
= $i_{0}\tau - i_{0}\tau + \frac{i_{0}\tau}{e}$
$q\ = \frac{i_{0}\tau}{e} = \frac{(E/R)(L/R)}{e} = \frac{el}{eR^{2}}$