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$$25\text{ mL}$$ of the given $$\text{HCl}$$ solution requires $$30\text{ mL}$$ of $$0.1\text{ M}$$ sodium carbonate solution. What is the volume of this $$\text{HCl}$$ solution required to titrate $$30\text{ mL}$$ of $$0.2\text{ M}$$ aqueous $$\text{NaOH}$$ solution?
$$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$$. The carbonate is $$30 \times 0.1 = 3$$ mmol, needing $$6$$ mmol of $$\text{HCl}$$, so $$[\text{HCl}] = \frac{6}{25} = 0.24\text{ M}$$. The $$\text{NaOH}$$ is $$30 \times 0.2 = 6$$ mmol, which needs $$\frac{6}{0.24} = 25\text{ mL}$$ of the acid.
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