What would be the molecule FeO, using the Roman Numeral method?

$\text{Ferrous oxide}$ $=$ $\text{Iron(II) oxide}$
We specify the oxidation state of the metal with Roman numerals. Oxygen generally has an oxidation state of $- I I$, and it does here. The sum of the oxidation numbers equals the charge on the species, and thus iron has an oxidation number of $+ I I$, i.e. $\text{ferrous oxide}$.
On the other hand, $\text{ferric oxide}$ is $F {e}_{2} {O}_{3}$, or $\text{Iron(III) oxide}$. The $\text{ous/ic}$ names are now a bit old-fashioned, but you will still see it the literature.