On addition of which metal, copper sulphate solution (blue colour) will be changed to colourless solution?

The tendency of an element to produce positive ions increases with a decrease in its standard reduction potential value. The solution will become colourless when Cu is reduced from it. So, the element needs to have a standard reduction potential that is lower than that of copper $\mathrm{Cu}$ in order to reduce the copper ion ${\mathrm{Cu}}^{2+}$ (cuprous) from its sulphate salt. 

Our knowledge of copper's typical reduction potential is as follows-
$\mathrm{E}°({\mathrm{Cu}}^{2+}/\mathrm{Cu})=0.34 \mathrm{V}$
Additionally, the following are the provided metals' typical reduction potentials-
$\mathrm{E}°({\mathrm{Fe}}^{2+}/\mathrm{Fe})=-0.44 \mathrm{V}$
$\mathrm{E}°({\mathrm{Ag}}^{+}/\mathrm{Ag})=0.80 \mathrm{V}$
$\mathrm{E}°({\mathrm{Zn}}^{2+}/\mathrm{Zn})=-0.76 \mathrm{V}$
$\mathrm{E}°({\mathrm{Hg}}^{2+}/{\mathrm{Hg}}_2^{2+})=0.92 \mathrm{V}$
The standard reduction potential values of the metals show that iron and zinc are stronger reducing agents than copper, whereas mercury and silver are weaker reducing agents than copper. Both of them namely, iron and zinc are stronger reducing agents than copper. As a result, copper can be removed from copper sulphate solution by iron and zinc.
Zinc will create ZnSO₄ while iron will produce FeSO₄.
$\mathrm{Zn}+{\mathrm{CuSO}}_4\to {\mathrm{ZnSO}}_4+\mathrm{Cu}$
$\mathrm{Fe}+{\mathrm{CuSO}}_4\to {\mathrm{FeSO}}_4+\mathrm{Cu}$
Both $\mathrm{Fe}$ and $\mathrm{Zn}$ will displace copper but only zinc sulphate produces a colourless solution, whereas, ferrous sulphate, which has coloured crystals, produces a green-coloured solution.

Hence, the correct answer is option C.