What volume of 0.20M Ca(OH)2 will neutralize 45.0 mL of a 1M solution of HClO3?

Well, the #"Ca"("OH")_2# will not all dissolve... so your question is problematic.

If you add #"HClO"_3# into #"Ca"("OH")_2# while mixing THOROUGHLY so that extra solid dissolves, then this problem may be more possible by Le Chatelier's principle (basic salts dissolve better in acidic solution), but if you add #"0.20 M"# #"Ca"("OH")_2# into #"HClO"_3#, that requires that the base FULLY dissolved first...

And so if you grab such a SATURATED solution, you only get #"0.011 M"#, and extra solid sinks to the bottom of the container you sourced it from.

The #K_(sp)# of #"Ca"("OH")_2# is #5.5 xx 10^(-6)#, whose mass action expression is

#5.5 xx 10^(-6) = ["Ca"^(2+)]["OH"^(-)]^2 = s(2s)^2 = 4s^3#

So, the solubility #s# is:

#s = ((5.5xx10^(-6))/4)^(1//3) = "0.011 M"#

Thus, the maximum concentration you can have is #"0.011 M"#, and the rest is well, a SOLID. The reaction is:

#"Ca"("OH")_2(aq) + 2"HClO"_3(aq) -> 2"H"_2"O"(l) + "Ca"("ClO"_3)_2(aq)#

In that volume of #"HClO"_3# you have:

#"1 M" cdot "0.045 L" = "0.045 mols HClO"_3 = "0.045 mols H"^(+)#

which react with

#0.045 cancel("mols OH"^(-)) xx ("1 mol Ca"("OH")_2)/cancel("2 mols OH"^(-)) = "0.0225 mols Ca"("OH")_2#

Therefore, in #"0.011 M"# of it, you would need

#"1 L"/(0.011 cancel"mol") xx 0.0225 cancel"mols" = color(blue)(ul"2.05 L")#

It sounds absolutely absurd, but #"Ca"("OH")_2# is NOT that strong of a strong base...

And in this #"2.05 L"#, there would have been #"28.71 g"# of #"Ca"("OH")_2(s)# that did NOT dissolve, that then falls back into the bottom of the container you sourced it from.