$4Mg+10{HNO}_3⟶4Mg{\left({NO}_3\right)}_2+{NH}_4{NO}_3+3H_{2}O$

The amount of magnesium required to reduce 1 mole of ${HNO}_3$ in the reaction is

The oxidation state of nitrogen in ${HNO}_3$ is $+5$

The oxidation state of nitrogen in $Mg{\left({NO}_3\right)}_2$ is $+5$

The oxidation state of nitrogen in ${NH}_4{NO}_3$ is $-3$ and $+5$

Only one nitrogen out of ten gets reduced

Four moles of magnesium reduces one mole of ${HNO}_3$

Atomic weight of magnesium is 24

Therefore, $4\times 24=1$ mole nitric acid

$96=1$ mole nitric acid

Hence the correct option is A.