The amount of $Mg$ in gms to be dissolved in dilute $H_2{SO}_4$ to liberate $H_2$ which is just sufficient to reduce $160 g$ of ferric oxide is

$F{e}_2{O}_3+3H_2\to 2Fe + 3H_{2}O$

Here, 160g of ${Fe}_2{O}_3=3\times 2=6 g$ of $H_2$

So, according to the question,

$Mg+H_{2}S{O}_4\to MgS{O}_4+H_2$

$24g of Mg =2g of H_2$

$So, 6g of H_2 will give = 24\times 6/2=72g of Mg$

Therefore, option (C) is correct.