The question talks about reaction between Fe and $CuS{O}_4$. Observe that weights of the reactants are given and the weight of a product is asked. Thus, it is a weight-weight stoichiometry problem.
In all such stoichiometry problems, start by balancing the reaction
Fe + $CuS{O}_4$ $\to$ $FeS{O}_4$ + Cu
From the equation, 1 mole of Fe = 1 mole of Cu
$\therefore$ 55.85g $\approx$ 63.6g of Cu
∴ 10g of Fe will displace 11.39g of Cu if available.
The solution has 70g of $CuS{O}_4$
1 mole $CuS{O}_4$ has 63.6g of Cu
i.e., [63.6 + 32 + 16(4)]g of $CuS{O}_4$ = 63.6g of Cu
159.6g of $CuS{O}_4$= 63.6g of Cu
70g of $CuS{O}_4$= 27.76g of Cu

$\therefore$  10 g = 10 $\times$ $\frac{63.6}{55.85}$ = 11.39 g of Cu

$\therefore$ In the $CuS{O}_4$ solution, 27.76g of Cu is present of which 11.39g of Cu will be precipitated out by the addition of 10g of Fe.