$?molNaOH=20gNaOH\times \frac{1molNaOH}{40gNaOH}=0/5molNaOH$
$\left[ NaOH \right]=\left[ O{{H}^{-}} \right]=\frac{0/5mol}{2L}=0/25mol.{{L}^{-1}}$
$\left[ O{{H}^{-}} \right]\left[ {{H}^{+}} \right]={{10}^{-14}}\Rightarrow 0/25\left[ {{H}^{+}} \right]={{10}^{-14}}\Rightarrow \left[ {{H}^{+}} \right]=4\times {{10}^{14}}mol.{{L}^{-1}}$
$pH=-log\left[ {{H}^{+}} \right]\Rightarrow pH=-log4\times {{10}^{-14}}=-(log4+log{{10}^{-14}})\Rightarrow pH=-(2\log 2-14)=-(0/6-14)=13/4$