If $1\frac{1}{2}$  moles of oxygen combine with Al to form ${\mathrm{Al}}_2{\mathrm{O}}_3,$ the weight of AI used in the reaction is: $(\mathrm{AI}=27)$

The molar ratio of $\mathrm{Al}:\mathrm{O}$ in ${\mathrm{Al}}_2{\mathrm{O}}_3,$ is

Therefore, Mole of $\mathrm{Al}=\frac{2}{3}\times \text{ Moles of }{\mathrm{O}}_2$

$\begin{array}{r}=\frac{2}{3}\times \frac{3}{2}\times 2\\ =2\text{ moles }\mathrm{Al}\end{array}$ 

Mass of $\mathrm{Al}=2\times 27=54\mathrm{g}$

So, option B is correct answer