A capacitor is connected to a 20V battery through a resistance of...

$V = {V}_{0}\left(1 - {e}^{- t / RC}\right)$

$2 = 20\left(1 - {e}^{- \dfrac{1\mus}{10C}}\right)$

$\dfrac{1}{10} = \left(1 - {e}^{- \dfrac{1 \times {10}^{- 6}}{10C}}\right)$

${e}^{- \dfrac{1 \times {10}^{- 6}}{10C}} = \dfrac{9}{10}$

$\dfrac{1 \times {10}^{- 6}}{10C} = \ln \left(\dfrac{10}{9}\right)$

$C = \dfrac{1 \times {10}^{- 6}}{10 \times \ln \left(\dfrac{10}{9}\right)} = \dfrac{{10}^{- 7}}{1 . 05} = \dfrac{1}{1 . 05}\mu F = \dfrac{100}{105}\mu F = 0 . 95\mu F$