$Fe_2O_3+Zn\rightarrow Fe^{2+}\rightarrow 100ml$ solution
25ml sample=17ml of 0.0167M of an oxidant.
Let 'n' be the no of electrons taken up by oxidant.
Now Meq of $Fe^{2+}$ in25ml=Meq of oxidant
$\Rightarrow [0.0167\times n]\times 17$
Meq of $Fe^{2+}$ in 100ml=$[0.0167\times n\times 17]\times 4$
Also meq of $Fe_2O_3$=Meq of $Fe^{2+}$ in 100ml
$\Rightarrow 0.0167\times 68]n$
Meq of $Fe_2O_3=\large\frac{0.552}{E}$$\times 1000$
$\Rightarrow (0.0167\times 68)n$
$E_{\large Fe_2O_3}=?$
1 mole of $Fe_2O_3=2Fe^{3+};Fe^{3+}+1e^-\rightarrow Fe^{2+}$
$X=2$ for 1 mole of $Fe_2O_3$
$\Rightarrow E_{Fe_2O_3}=\large\frac{160}{2}$$=80$
$n=\big[\large\frac{0.552}{80}$$\times 1000\big]/[0.0167\times 68]=6$
$\Rightarrow$ Electrons taken by Oxidant=6
Hence (d) is the correct answer.