Calculate the moles of Na₂S₂O₃ used in the titration (t₂).

The mean titre (t₂) = 30.55 cm³ = 0.03055 dm³
The concentration of Na₂S₂O₃ (from part a) = 0.3 mol/dm³

$$\text{Moles of Na}_2\text{S}_2\text{O}_3 = \text{Concentration} \times \text{Volume} = 0.3 \text{ mol/dm}^3 \times 0.03055 \text{ dm}^3 = 0.009165 \text{ mol}$$

The reaction is: I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)
From the stoichiometry, 1 mole of I₂ reacts with 2 moles of S₂O₃²⁻.

$$\text{Moles of I}_2 = \frac{1}{2} \times \text{Moles of Na}_2\text{S}_2\text{O}_3 = \frac{1}{2} \times 0.009165 \text{ mol} = 0.0045825 \text{ mol}$$

The reaction is: Cr₂O₇²⁻(aq) + 6I⁻(aq) + 14H⁺(aq) → 3I₂(aq) + 2Cr³⁺(aq) + 7H₂O(l)
From the stoichiometry, 1 mole of Cr₂O₇²⁻ produces 3 moles of I₂.

$$\text{Moles of Cr}_2\text{O}_7^{2-} = \frac{1}{3} \times \text{Moles of I}_2 = \frac{1}{3} \times 0.0045825 \text{ mol} = 0.0015275 \text{ mol}$$

The volume of solution M used for titration was 25 cm³ = 0.025 dm³.

$$\text{Concentration of Cr}_2\text{O}_7^{2-} = \frac{\text{Moles}}{\text{Volume}} = \frac{0.0015275 \text{ mol}}{0.025 \text{ dm}^3} = 0.0611 \text{ mol/dm}^3$$

The concentration of dichromate ions, Cr₂O₇²⁻, in solution M is $\boxed{\text{0.0611 mol/dm}^3}$.