For the element fluorine, there is only one stable isotope, \(^{19}_{9}\text{F}\), which has an abundance of 100%.
The relative atomic mass (A_r) of fluorine in this case is simply the mass of the isotope, which is approximately 19. The relative atomic mass to three significant figures is:
\[ \text{A_r (Fluorine)} = 19.0 \]
So the answer is:
Relative atomic mass (A_r) to (3sf) = 19.0