BaCl2 + H2SO4 ==> BaSO4 + 2HCl
BaCl2 = 0.1 M x 0.1 L = 0.01 mols
H2SO4 = 0.05 M x 0.1 L = 0.005 mols.
Therefore, all of the H2SO4 will react, 1/2 of the BaCl2 will react.
Mols BaCl2 at end of reaction = 0.01 - 0.005 = 0.005 mols.
Mols H2SO4 at end of reaction = 0.005 - 0.005 = 0
Mols BaSO4 formed = 0.005.
Mols HCl formed = 2 x 0.005 = 0.01.
So you have two sources for Ba^+2. One is the mols Ba^+2 from BaCl2 and the other is any Ba^+2 from BaSO4. But BaSO4 is quite insoluble and the amount of Ba^+2 contributed by BaSO4 is quite small and we can neglect it. So Ba^+2 from BaCl2 is simply mols/L = 0.005/total volume. total volume is 100 mL + 100 mL or 200 mL = 0.2 L. You do the math.