To find the molarity $c$, you need to know the amount $n$ required for a titration:
$$c(\ce{Na2S2O3}) = \frac{n(\ce{Na2S2O3})}{V(\ce{Na2S2O3})}$$
Unknown amount $n(\ce{Na2S2O3})$ can be found from the second balanced reaction:
$$\ce{I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6}$$
$$n(\ce{Na2S2O3}) = 2\cdot n(\ce{I2})$$
Finding $n(\ce{I2})$ is trivial if you would've written the first redox equation correctly, and that's where the problem in your solution arises.
The incomplete first equation resulting in wrong stoichiometry and false ratio between $\ce{KIO3}$ and $\ce{I2}$.
The reaction takes place in acidic media (I added [diprotic] sulfuric acid, but basicity of acid doesn't matter), and the ratio between $\ce{KIO3}$ and $\ce{I2}$ won't be $1:1$:1
$$\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$$
$$n(\ce{I2}) = 3\cdot n(\ce{KIO3}) = \frac{3\cdot m(\ce{KIO3})}{M(\ce{KIO3})}$$
With all the steps sorted out, the final expression for the molarity of hyposulfite can be rewritten algebraically and solved by plugging in the values:
$$
\begin{align}
c(\ce{Na2S2O3}) &= \frac{2\cdot 3\cdot m(\ce{KIO3})}{M(\ce{KIO3})\cdot V({\ce{Na2S2O3})}}\\
&= \frac{2\times 3\times \pu{0.1 g}}{\pu{214.0 g mol-1}\times\pu{4.4E-2 L}} \\
&\approx \pu{6.4E-2 mol L-1}
\end{align}
$$
1 All you need to do to prove that is to write half-reactions for the redox process between iodate and iodide:
$$
\begin{align}
\ce{2\overset{+5}{I}O3- + 12 H+ + 10 e- &-> \overset{0}{I}_2 + 6 H2O} & \tag{red}\\
\ce{2\overset{-1}{I}^- &-> \overset{0}{I}_2 + 2 e-} &|\cdot 5 \tag{ox}\\
\hline
\ce{2 IO3- + 10I- + 12 H+ &-> 6 I2 + 6 H2O} \tag{redox}
\end{align}
$$
or, after dividing the total redox reaction by 2, finally
$$\ce{IO3- + 5I- + 6 H+ -> 3 I2 + 3 H2O}$$