Deposition of saturated solution and affect of evaporated water
Unfortunately, your calculations are a little bit wrong because of the poor choice of the total initial mass. It is $200 g$, and not $220~ g$. So here are the right calculations.
As $100 g$ water dissolves $109g$ $\ce{KNO3}$ at $60$°C, producing a total mass of $209g$, you may deduce that $200g$ of your initial solution contains $(200/209)·109g = 104.3 g$ $\ce{KNO3}$ and $(200/209)·100g = 95.7g$ water.
Now you loose $20 g$ water by evaporation. The amount of $\ce{KNO3}$ is still $104.3 g$. Remaining water is $95.7g - 20g = 75.7g$ water. At the final temperature $20$°C, $100g$ water can only dissolve $32g$ $\ce{KNO3}$. As a consequence, $75.7$ g water can dissolve $(75.7/100)·32g = 24.2 g$ $\ce{KNO3}$. The rest is deposited in the flask. It is $104.3 g - 24.2 g = 80.1 g$.
Please note that the mass has a unit, the gram. It is incorrect to say &quot;the solubility of $\ce{KNO3}$ in $100g$ of water is $32$ at $20°$C and $109$ at $60°$C&quot;. You should have said that &quot;the solubility of $\ce{KNO3}$ in $100$g of water is $32 g$ at $20$°C and $109$ g at $60$°C