We want to find the mol of Ba2+ that reacted in the 10ml volumetric flask.
To do this we need to calculate the initial mass in this 10 ml flask.
m(Ba2+)= 200mg but this is in 1L(1000ml), since we have only 2ml we need to adjust this by multiplying 200mg by (2/1000).
We get m(Ba2+)= 200mg X (2/1000) = 0.400 mg in 10ml volumetric flask(amount initially)
now we figure the mass of Ba2+ that has reacted.
From the graph,and from part a, we figured that the m(Ba2+) remaining or in other words, is in excess, is 0.19mg.
So to figure the mass of Ba2+ that reacted to produce the precipitate, we simply subtract mass excess from mass of initial.
mass(Ba2+)reacted= initial-excess=0.400-0.19=0.21mg
now to find mol of Ba2+ reacted.
n(Ba2+)= (0.21X10
-3)/137.3=1.53X10
-6mol (2 decimal places because of the subtraction in the above calculation)
Hope that helps

Edit: just realised you were just asking for the m of Ba2+ added to volumetric flask...so yeh just refer to bolded bit above