b
During time $'{t_2}'$ capacitor is discharging with the help of resistor $'R'$
$\therefore {\rm{q}} = {{\rm{q}}_0}{{\rm{e}}^{ - {\rm{t/RC}}}}$
${\rm{V}} = {{\rm{V}}_0}{{\rm{e}}^{ - t/{\rm{RC}}}}$ $[\because \mathrm{Q}=\mathrm{CV}]$
As $\mathrm{V}_{0}=\frac{2 \mathrm{V}}{3} ; \mathrm{V}=\frac{\mathrm{V}}{3}$
${t_2} = RC\,\ell n\,2$
During time ${{\rm{t}}_1}{\rm{' }}$ capacitor is charging with the help of battery.
$\therefore {\rm{q}} = {{\rm{q}}_0}\left( {1 - {{\rm{e}}^{ - t/{\rm{RC}}}}} \right)$ or ${\rm{V}} = {{\rm{V}}_0}\left( {1 - {{\rm{e}}^{ - t/{\rm{RC}}}}} \right)$
as $\mathrm{V}_{0}=\frac{2 \mathrm{V}}{3} ; \mathrm{V}=\frac{\mathrm{V}}{3}$
${{\rm{t}}_1} = {\rm{RC}}\,\ell n\,2$
${\rm{T}} = {{\rm{t}}_1} + {{\rm{t}}_2} = 2{\rm{RC}}\,\ell n\,2$