Rabigh's answer is very good, I just want to add an alternative approach to using the dilution calculations equation, #C_1V_1 = C_2V_2#.
You're dealing with #"100 cm"^3# of a 0.1-M #NaOH# solution. Because sodium hydroxide is a strong base, it will dissociate completely in aqueous solution to give #Na^(+)# cations and #OH^(-)# anions.
#NaOH_((aq)) -> Na_((aq))^(+) + OH_((aq))^(-)#
You can use this solution's molarity to determine the number of moles of sodium hydroxide, which is equal to the number of moles of #OH^(-)# ions, you have in solution before adding water
#C = n/V => n = C * V#
#n_("NaOH") = "0.1 M" * 100 * 10^(-3)"L" = "0.01 moles NaOH"#
When you add the #"100 cm"^3# of water, the number of moles of sodium hydroxide remains unchanged; the only thing that changes is the volume of the solution, which implies that the concentration of #OH^(-)# will change as well.
Think of it like this: same number of moles, twice the volume #-># half the initial concentration
#C_("new") = n/V_("total") = "0.01 moles"/((100 + 100) * 10^(-3)"L") = "0.05 M"#
Use this concentration to determine pOH
#pOH = -log([OH^(-)]) = -log(0.05) = 1.3#
Therefore, pH is equal to
#pH_("sol") = 14 - pOH = 14 - 1.3 = color(green)(12.7)#
