What is the pH of 0.001 M NaOH(aq) at 298 K?

The problem provides you with the temperature of the solution because at #25^@"C"#, which is about #"298 K"#, an aqueous solution has

#color(blue)(ul(color(black)("pH + pOH = 14")))#

This means that you can express the #"pH"# of the solution in terms of its #"pOH"#, which, as you know, is defined as

#"pOH" = - log(["OH"^(-)])#

You can thus say that the #"pH"# of the solution is equal to

#"pH" = 14 - "pOH"#

#"pH" = 14 - [ - log(["OH"^(-)])]#

#"pH" = 14 + log(["OH"^(-)])#

So instead of calculating the #"pH"# of the solution by using the concentration of hydronium cations, #"H"_3"O"^(+)#

#"pH" = - log(["H"_3"O"^(+)])#

you can do so indirectly by using the concentration of hydroxide anions.

Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to you here, and hydroxide anions.

Both ions are produced in #1:1# mole ratio with the strong base, so you can say that

#["OH"^(-)] = ["NaOH"] = "0.001 M"#

Now that you know the concentration of hydroxide anions in this solution, you can say that its #"pH"# is equal to

#"pH" = 14 + log(0.001) = color(darkgreen)(ul(color(black)("11.0")))#

The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the strong base.