The correct option is A 0.328 g
So the given solution will form acidic buffer for which pH is given by
pH=pKa+log[salt][acid]
6=5+log[salt][acid]
∴log[salt][acid]=1
[salt][acid]=10
initially 10 mmoles each of salt and acid was present when x mmoles of NaOH is added
(10+x)(10−x)=10
x=8.18 mmoles
wNaOH=40×8.18×10−3=0.327g