An inductor of 10mH is connected to a 20V battery through a...

{I}_\max = \dfrac{V}{R} = \dfrac{20V}{10KΩ} = 2mA

For $LR$-decay circuit,

I = {I}_\max{e}^{- Rt / L}

$I = 2mAe\dfrac{- 10 \times {10}^{3} \times 1 \times {10}^{- 6}}{10 \times {10}^{- 3}}$

$I = 2mA{e}^{- 1}$

$I = 2 \times 0 . 37mA$

$I = \dfrac{74}{100}mA$

$x = 74$