The statement is false.
Work Step by Step
To find out which one has a larger first ionization energy, we must compare the effective nuclear charge on the outermost electrons of both element to find out which outermost electron undergoes larger attraction from the nucleus. $O$ has a nuclear charge of $8+$ and has 2 inner electrons and 6 outer electrons. $F$ has a nuclear charge of $9+$ and has 2 inner electrons and 7 outer electrons. $F$ obviously has a higher nuclear charge than $O$. In addition, both have the same number of inner electrons. Though $F$ also has more outer electrons than $O$, these outer electrons are not quite effective in shielding the attraction from the nucleus. Therefore, the outer electrons in $F$ experiences a higher attraction from the nucleus than those in $O$ and as a result, harder to be removed. Therefore, the first ionization energy of $F$ would be higher than that of $O$. The statement is false. (Another way to look at this is through the radii of the atoms. $F$ has a smaller radium than $O$, since there is more attraction to the outer electrons from the nucleus, pulling them closer to the nucleus. That means the first ionization energy of $F$ must be higher.)