Following reaction takes place when $\mathrm{Al}$reacts with $\mathrm{NaOH}$.

$2\mathrm{Al}\stackrel{\mathrm{NaOH}}{\to }2\mathrm{A}(\mathrm{OH}{)}_3\stackrel{∆}{\to }{\mathrm{Al}}_2{\mathrm{O}}_3$

$2$ moles of $\mathrm{Al} (54 \mathrm{g})$ react with $\mathrm{NaOH}$to produce $1 \mathrm{mole}$ of ${\mathrm{Al}}_2{\mathrm{O}}_3$ $(102 \mathrm{g})$.

So,$1$ $(27 g)$mole of $\mathrm{Al}$ reacts with$\mathrm{NaOH}$ to produce $51 \mathrm{g}$ of ${\mathrm{Al}}_2{\mathrm{O}}_3$.