There are a couple of things wrong with your question
- you know that the solution's molarity is 2.05 M, so I presume that you're interested in its molality;
- the density of acetic acid solutions doesn't come close to 1.20 g/mL (pure acetic acid has a density of 1.05 g/mL), so I presume that you meant to write 1.02 g/mL.
Let's assume a 1-L sample of your solution. This solution would contain
In order to get the mass of water, you need to use the solution's density.
Since the solution contains 2.05 moles of acetic acid, you can use its molar mass to determine how many grams of acetic acid you have.
This measn that the solution will contain
This means that the solution's molality will be
Rounded to three sig figs, the answer will be