X forms an oxide $X_2{O}_3$, $0.36$ grams of X forms $0.56$ grams of $X_2{O}_3$. So, the atomic weight of X is

The atomic mass of oxygen is $16$ g. Let the atomic mass of X be y g.

Molecular mass of X₂O₃=$2$(atomic mass of X) +$3$(atomic mass of oxygen)

                                          =$2$(y g) + $3$(16)

                                         =$2$y + $48$ grams

$2$y g of X forms ($2$y + $48$) g of $X_2{O}_3$

By using unitary method,

Let, $0.36$ g of X forms ($2$y+$48$)/$2$y × $0.36$ g, which is $0.56$g of $X_2{O}_3$

(($2$y+$48$)/$2$y ×$0.36$)) = $0.56$

(($2$y+$48$)/$2$y = $56/36$

On solving the above equation, we get the atomic mass of X is $43.2$g

Hence, the correct option is D.