We're asked to find the molarity of the #"NaOH"# solution given some titration values.

Let's first write the balanced chemical equation for this reaction:

#2"NaOH"(aq) + "H"_2"SO"_4(aq) rarr "Na"_2"SO"_4 (aq) + 2"H"_2"O" (l)#

We can calculate the number of moles of #"H"_2"SO"_4# used by using the molarity equation:

#"mol solute" = ("molarity")("L soln")#

We'll convert the volume to liters to use this equation:

#"mol H"_2"SO"_4 = (0.5"mol"/(cancel("L")))(0.0205cancel("L")) = color(red)(0.0103# #color(red)("mol H"_2"SO"_4#

Using the coefficients of the chemical equation, we can determine thee relative number of moles of #"NaOH"# used:

#color(red)(0.0103)cancel(color(red)("mol H"_2"SO"_4))((2color(white)(l)"mol NaOH")/(1cancel("mol H"_2"SO"_4))) = 0.0205# #"mol NaOH"#

Lastly, we'll use the given volume of the #"NaOH"# solution and the mole value to calculate the molarity of the solution:

#"molarity" = "mol solute"/"L soln"#

#= (0.0205color(white)(l)"mol NaOH")/(0.020color(white)(l)"L") = color(blue)(1M#

which I'll leave to one significant figure I suppose (real value #= 1.025M#).