How many fluorine atoms can be made from 21.3 g of boron trifluoride?

Molar mass of BF₃ = (10.8 + 19.0×3) g/mol = 67.8 g/mol
No. of moles of BF₃ = (21.3 g) / (67.8 g/mol) = 0.314 mol

Each moles of BF₃ contains 3 moles of F atoms.
No. of moles of F atoms = (0.314 mol) × 3 = 0.942 mol

Each mole of F atoms contains 6.022 × 10²³ F atoms.
No. of F atoms can be made = (0.942 mol) × (6.022 × 10²³ F atoms/mol) = 5.67 × 10²³ F atoms

How many atoms?

Simply use the unit-factor method and don't worry about intermediate answers.

21.3g BF3 x (1 mol BF3 / 67.81g BF3) x (3 mol F atoms / 1 mol BF4) x (6.022x10^23 atoms F / 1 mol atoms F) = 5.67x10^23 atoms F