\(2=20\left(1-{e}^{-t / {RC}}\right)\)
\(\frac{1}{10}=1-{e}^{-t / {RC}}\)
\({e}^{-t / R C}=\frac{9}{10}\)
\({e}^{{t} / {RC}}=\frac{10}{9}\)
\(\frac{{t}}{{RC}}=\ln \left(\frac{10}{9}\right) \Rightarrow {C}=\frac{{t}}{{R} \ell {n}\left(\frac{10}{9}\right)}\)
\({C}=\frac{10^{-6}}{10 \times .105}=.95\,\mu {F}\)