Supposing complete ionization:
BaBr2 → Ba{2+} + 2 Br{-}
(2.23 × 10^–4 g BaBr2) / (297.135 g BaBr2/mol) / (2.00 L) = 3.75 × 10^-7 mol/L BaBr2
(3.75 × 10^-7 mol/L BaBr2) x (1 mol Ba{2+} / 1 mol BaBr2) = 3.75 × 10^-7 mol/L Ba{2+}
(3.75 × 10^-7 mol/L BaBr2) x (2 mol Br(-} / 1 mol BaBr2) = 7.50 × 10^-7 mol/L Br{-}