First, let's write out the balanced equation, so we know what we're dealing with

4 Al + 3 O₂  ⇒  2 Al₂O₃


Now, we'll convert g O₂ into g Al
g to mole  ↔  mole to mole  ↔  mole to g

23.7 g O₂ 

(1 mole O₂) / (32.00 g O₂)

(4 moles Al) / (3 moles O₂)

(26.98 g Al) / (1 mole Al)

= 26.6 g Al

26.6 g Al are required to completely react with 23.7 g O₂