${10}^{-2}$ mole of NaOH was added to 10 liters of water. The pH will change by

${\mathrm{p}}^{\mathrm{H}}$of water before adding base = 7 because water is neutral.

New $\left[{\mathrm{OH}}^{-}\right] ={10}^{-2} \times \frac{1}{10} = {10}^{-3}$

New  pH = 11

Change in pH = 11 - 7 = 4