Molecular weight of haemoglobin $=67200(Given)$

$\%$ of Fe present in the sample $=0.33 \%$

$\therefore$ Amount of $Fe$ present in haemoglobin $=\large\frac{0.33 }{100}$$ \times 67200$

$\qquad= 222\;amu$

Thus the number of atoms present $=\large\frac{222}{56}$

$\qquad= 3.9 =4 \;amc$

Hence c is the correct answer.