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Molar mass of $K_2Cr_2O_7 = (2 \times 39.1) + (2 \times 52) + (7 \times 16) = 294.2\text{ g/mol}$

Concentration of $K_2Cr_2O_7$ solution,
$S_1 = \frac{w}{M \times V \text{ (in L)}} = \frac{1.47111}{294.2 \times 0.250} \approx 0.02\text{ M}$

The balanced redox reactions are:

1. $\mathrm{Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O}$
2. $2\mathrm{S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-}$

From the stoichiometric ratio:
$1\text{ mol } K_2Cr_2O_7 \equiv 3\text{ mol } I_2 \equiv 6\text{ mol } Na_2S_2O_3$

Applying the equivalence relation:
$n_{\mathrm{Na_2S_2O_3}} = 6 \times n_{\mathrm{K_2Cr_2O_7}}$
$V_2 \times S_2 = 6 \times V_1 \times S_1$

Given, $V_1 = 20\text{ mL}$, $V_2 = 20\text{ mL}$:
$20 \times S_2 = 6 \times 20 \times 0.02$
$S_2 = 6 \times 0.02 = 0.12\text{ M}$

Therefore, the concentration of the $Na_2S_2O_3$ solution is $0.12\text{ M}$.