If the alloy has m g of copper, then $(12.9-\mathrm{m})$ g of zinc will be present. Therefore, the volumes of copper and zinc will be,

${\mathrm{V}}_{\mathrm{Cu}}=\frac{\mathrm{m}}{{\mathrm{\rho }}_{\mathrm{Cu}}} \& {\mathrm{V}}_{\mathrm{Zn}}=\frac{12.9-\mathrm{m}}{{\mathrm{\rho }}_{\mathrm{Zn}}}$

Thus the total volume of alloy,

$\mathrm{V}={\mathrm{V}}_{\mathrm{Cu}}+{\mathrm{V}}_{\mathrm{Zn}}=\frac{\mathrm{m}}{8.9}+\frac{(12.9-\mathrm{m})}{7.1}\mathrm{\_\_\_\_\_}\left(1\right)$

12.9 - m g of zinc will be present in the alloy if it has m g of copper. 

The volumes of copper and zinc will be the amount of water displaced

$= (12.9 - 11.3) \mathrm{g} \mathrm{or} \mathrm{V} = 1.6 \mathrm{cc\_\_\_\_\_\_\_\_}\left(2\right)$

When we solve for m by substituting V from equation (ii) in I we obtain m = 7.61 g.

Hence the correct answer is 7.6gm.