I assume that the concentration of the Na₂S₂O₃ was 0.1 mol•dm⁻³.
The equations are:
BaCrO₄ → Ba²⁺ + CrO₄²⁻
2CrO₄²⁻ + 6I⁻ + 16H⁺ → 2Cr³⁺ + 3I₂ + 8H₂O
I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
Moles of I₂ = 0.030 dm³ S₂O₃²⁻ ×
Moles of CrO₄²⁻ =0.002 mol I₂ ×
Mass of BaCrO₄ = 0.001 mol CrO₄²⁻ ×
Mass of BaSO₄ = 0.929 g – 0.3 g = 0.7 g
Moles of SO₄²⁻ = 0.7 g BaSO₄ ×
Note: I have calculated the answer to only 1 significant figure. That is all you gave me for the concentration of the Na₂S₂O₃. If you need more precision, you will have to recalculate.