Step 1 — Find molarity of HCl:
$\text{Na}_2\text{CO}_3 + 2\text{HCl} \to 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$
Moles Na$_2$CO$_3 = 0.030 \times 0.1 = 3\times10^{-3}\ \text{mol}$
Moles HCl needed $= 2 \times 3\times10^{-3} = 6\times10^{-3}\ \text{mol}$
$M_{\text{HCl}} = \dfrac{6\times10^{-3}}{0.025} = 0.24\ \text{M}$

Step 2 — Volume for NaOH:
$\text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O}$
Moles NaOH $= 0.030 \times 0.2 = 6\times10^{-3}\ \text{mol}$
$V_{\text{HCl}} = \dfrac{6\times10^{-3}}{0.24} = 0.025\ \text{L} = \mathbf{25\ \text{mL}}$