Number of atoms in 558.5g $Fe$ (at.wt =55.85) is

$\begin{array}{1 1}(a)\;Twice\;that\;in\;60g\;carbon\\(b)\;6.023\times 10^{22}\\(c)\;Half\;of\;8g\;He\\(d)\;558.5\times 6.023\times 10^{23}\end{array}$

Moles of Fe =$\large\frac{558.5}{55.85}$
$\Rightarrow 10$
Moles of C=$\large\frac{60}{12}$
$\Rightarrow 5$
$\therefore$ Number of atoms in 558.5g is twice that in 60g C
Hence (a) is the correct answer.