Consider the LR circuit shown in the figure. If the switch S is...

The current in the LR circuit at any time $t$ is
$I = \dfrac{E}{R}\left(1 - {e}^{- \dfrac{Rt}{L}}\right)$

Where $E, R, L, I$ are the EMF, resistance, inductance and the current respectively.
$∴$ Charge flown through the circuit from time 0 to $\dfrac{L}{R}$ is
(charge) $∆ q = \mathop{∫}_{0}^{\dfrac{L}{R}}I . dt = \mathop{∫}_{0}^{\dfrac{L}{R}}\dfrac{E}{R}\left(1 - {e}^{- \dfrac{Rt}{L}}\right)dt$
$⇒ ∆ q = \dfrac{E}{R}{\left[t + \dfrac{L}{R}{e}^{- \dfrac{Rt}{L}}\right]}_{0}^{\dfrac{L}{R}}$
$⇒ ∆ q = \dfrac{E}{R}\left[\dfrac{L}{R} + \dfrac{L}{R}{e}^{- 1} - \dfrac{L}{R}\right]$
$⇒ ∆ q = \dfrac{E}{R} \times \dfrac{L}{eR}$
$⇒ ∆ q = \dfrac{EL}{e{R}^{2}} = \dfrac{EL}{2.7{R}^{2}}$