Explanation

(b)
100 mL of 0.01 M $NaOH$ solution is diluted to $1\ dm^{3}$ ($i.e$., 10 times diluted hence, the resultant solution will be 0.001 M)
$\left\lbrack OH^{-} \right\rbrack = 0.001 = 10^{- 3}$
$\left\lbrack H^{+} \right\rbrack = \frac{10^{- 14}}{\lbrack OH^{-}\rbrack} = \frac{10^{- 14}}{10^{- 3}} = 10^{- 11}\$
$pH = - \log{\lbrack H^{+}\rbrack}$
$= - \log{\lbrack 10^{- 11}\rbrack}$
pH=11