The amount of substance of the solute #n# in a solution is
#color(red)(n=cV)#, where
#c#: molarity (mol/L)
#V#: volume of the solution (L)
Then, let #v#(mL) the volume of #0.5# M calcium carbonate solution.
The amount of substance of #CaCO_3# in #v# ml of #0.5# M #CaCO_3# solution is #0.5 (M) * v/1000 (L) = 5.0xx10^-4 v# mol. [1]
And, #200# ml of #0.2# M #CaCO_3# solution contains #0.2*200/1000# = #4.0xx10^-2# mol of calcium carbonate. [2]
[1] and [2] must be same, since the amount of substance of #CaCO_3# doesn't change after dilution.
Thus,
#5.0xx10^-4 v= 4.0xx10^-2#
is the equation to solve. #v=80# is the answer.
You need #80# ml of stock solution and #200-80=120# ml of pure water.
