Finding molarity of hyposulfite solution used in titration
Your method in using the equivalent concept is correct, but the miscorrect in your solution is in determining the n-factor.

Refer to the above equations and the calculation N-factorPoutnik
N-factor Is the change of atom oxidation state multiplied by the number of atoms changing this state.
Because the iodine is just an intermediate product to be titrated by $\ce{Na2S2O3}$, the oxidation number change is by  $6$.Therefore, n-factor of  $\ce{KIO3}$ is $6$.So, the equivalent mass is M/6.And , n-factor of  $\ce{Na2S2O3}$ is $1$.So, the equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass.
$$
\begin{array}{cc}
\ce{&amp;KIO3 &amp;+ 5KI + 3H2SO4 -&gt; &amp;3I2 +&amp;3K2SO4 + 3H2O}\\
&amp;\frac{0.6}{214}~\mathrm{meq} &amp; &amp;\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$
\begin{array}{cc}
\ce{&amp;I2 &amp;+ &amp;2 Na2S2O3 &amp;-&gt; 2 NaI + Na2S4O6}\\
&amp;\frac{0.6}{214}~\mathrm{meq} &amp; &amp;\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$∴\frac{0.6}{214} = 44\times10^{-3}\times x\times 1 \to n\text{-factor of hypo}$$
$$∴x \approx{\pu{0.064 M}}$$