Answer
$m(Fe) = 675g$
Work Step by Step
The initial mass of the ore is $2500g$. The mass of magnetic iron oxide, $Fe_3O_4$, in this ore equals to $m(Fe_3O_4) = 0.432\cdot 2500g = 1080g$.
Simplified reduction reaction is as follows:
$Fe_3O_4 + 4CO \rightarrow 3Fe + 4CO_2$
In theory, the amount of the iron obtained by this reaction is three times larger than the amount of $Fe_3O_4$ reacted, therefore:
$n(Fe) = 3\cdot n(Fe_3O_4) = 3\cdot \frac{m(Fe_3O_4)}{M(Fe_3O_4)} = 3\cdot \frac{1080g}{232\frac{g}{mol}}\approx 14mol$
Hence, $m(Fe) = n(Fe) \cdot A(Fe) = 14mol \cdot 56\frac{g}{mol} = 782g$.
Only $86.3\%$ of this mass of iron can be recovered, so the mass of recovered iron equals $0.863\cdot 782g \approx 675g$.
