To calculate the molarity of the sodium thiosulphate solution, we'll need to apply stoichiometry principles. First, we must determine the number of moles of $ \text{KIO}_3 $ as it will relate to the number of moles of $ \text{I}_2 $ liberated during the reaction. Then we will use the volume of the sodium thiosulphate solution to find its molarity.

The reaction for the iodometry where $ \text{I}_2 $ is liberated from $ \text{KIO}_3 $ and reacted with the excess of $ \text{KI} $ is:

$ \text{KIO}_3 + 5 \text{KI} + 6 \text{HCl} \rightarrow 3 \text{I}_2 + 6 \text{KCl} + 3 \text{H}_2\text{O} $

From the balanced equation, we see that 1 mole of $ \text{KIO}_3 $ produces 3 moles of $ \text{I}_2 $.

Now let's find the number of moles of $ \text{KIO}_3 $:

$ \text{Moles of } \text{KIO}_3 = \frac{\text{mass of } \text{KIO}_3}{\text{molar mass of } \text{KIO}_3} $

$ = \frac{0.10 \text{ g}}{214.0 \text{ g/mol}} $

To get the number of moles, we need to perform the division:

$ = \frac{0.10}{214.0} $

$ = 4.6728972 \times 10^{-4} \text{ moles of } \text{KIO}_3 $

Since 1 mole of $ \text{KIO}_3 $ yields 3 moles of $ \text{I}_2 $, we multiply the moles of $ \text{KIO}_3 $ by 3:

$ \text{Moles of } \text{I}_2 = 3 \times 4.6728972 \times 10^{-4} $

$ = 1.4018692 \times 10^{-3} \text{ moles of } \text{I}_2 $

Next, $ \text{I}_2 $ reacts with sodium thiosulphate ($ \text{Na}_2\text{S}_2\text{O}_3 $) in the following stoichiometric reaction:

$ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 $

From the reaction, 1 mole of $ \text{I}_2 $ reacts with 2 moles of sodium thiosulphate. Therefore, the moles of sodium thiosulphate required to react with the $ \text{I}_2 $ produced can be calculated as follows:

$ \text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = 2 \times \text{Moles of } \text{I}_2 $

$ = 2 \times 1.4018692 \times 10^{-3} $

$ = 2.8037384 \times 10^{-3} \text{ moles of } \text{Na}_2\text{S}_2\text{O}_3 $

Now we have to calculate the molarity of the sodium thiosulphate solution. Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:

$ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} $

We have the number of moles of sodium thiosulphate and the volume is given as 45.0 mL which needs to be converted to liters:

$ \text{Volume in liters} = 45.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.045 \text{ L} $

Substitute the number of moles and the volume into the molarity formula:

$ M = \frac{2.8037384 \times 10^{-3} \text{ moles}}{0.045 \text{ L}} $

Performing the division gives us the molarity:

$ M = \frac{2.8037384 \times 10^{-3}}{0.045} $

$ = 0.0623053 \text{ M} $

So, the molarity of the sodium thiosulphate solution is approximately $ 0.0623 $ M.