Optical purity

14% of R racemise the 14% of S leaving 72% as it is !

%EE = 72 % S (+)
Racemic = 100 -72 = 28 % (50:50 mixture of enantiomers) ½ Racemic is S(+) lactic acid and ½ Racemic is R(-)-lactic acid. Total S (+) = 72 + 14 = 86% Total R (-) = 14

An optical purity of $72\%$ means that S-isomer is $72\%$ in excess of R-isomer, therefore, the remaining $(100\%-72\%=) 28\%$ of the mixture is evenly shared by R- and S-isomer. This means that R-isomer is $(28\%/2 =) \boxed{14\%}$ .