In the formation of $A{l}_2{O}_3$ from $Al$ and $O_2$, if$1.5$ mole of oxygen is used up, the mass of aluminum that reacted is 

The balanced equation is: 

$4Al+3O_2\to 2A{l}_2{O}_3$

From the equation we can conclude that 

$3$ moles of oxygen reacted with $4$ moles of aluminum

Then $1$  moles of oxygen reacted with $4/3$ moles of aluminum

Then $1.5$ moles of oxygen reacted with $4/3\times 1.5$ moles of aluminium

Upon calculation we get $2$ moles of aluminium

Molecular weight of aluminium is $27g/mol$. Then $2$  moles of aluminium 

weight is $2\times 27=54g$

 Correct option is (B).