Answer
$8.80\%$
Work Step by Step
Number of moles of the oxide:
$127\ mg\div 101.961\ mg/mmol=1.255\ mmol$
From stoichiometry:
$1.255\ mmol\times 2_{Al(OH)_3}/1_{Al_2O_3}\times 1_{Al}/1_{Al(OH)_3}=2.491\ mmol$ of Al
Mass fraction:
$2.491\ mmol\times26.982\ mg/mmol=67.2\ mg$
$67.2/764\times100\%=8.80\%$
