To make 2 molecules of Fe2O3 "rust", six O atoms are needed, along with two iron atoms. That would require three water (H2O) molecules, if there were no other source of oxygen. Therefore the correct answer is False.
The issue is what is meant by rust. In the real world, there are two iron oxides:
FeO and Fe2O3 which are the common oxides for the two common valence forms of the iron molecule.
In common rust, these two oxides and their hydroxides are present, almost half and half, along with trapped water.
So we give "rust" mixture a "formula" (EVEN THOUGH IT IS NOT A CHEMICAL COMPOUND, but a mixture) Rust consists of hydrated iron(III) oxides Fe2O3·nH2O and iron(III) oxide-hydroxide FeO(OH)·Fe(OH)3. Again, the statement is false using this understanding of rust.
This leaves us with the IRONII oxide, FeO, which is quite rare. In that case, the statement is true, but this is not refered to as "rust" by anyone I know.