Answer
$99.9993\%$
Work Step by Step
Final concentration of Calcium ion:
$Ksp=[Ca^{2+}][CO_3^{2-}]$
$3.4\times10^{-9}=[Ca^{2+}]0.050$
$[Ca^{2+}]=6.8\times10^{-8}\ M$
Percentage of calcium removed:
$(0.01-6.8\times10^{-8})/0.01\times100\%=99.9993\%$
