If yield of third reaction is $90$ percent, then moles of $C{O}_2$ formed, when $2.06\times {10}^3$gm $NaBr$ is formed, is:

The correct option is B $10$
Balanced chemical equations are -

$Fe+B{r}_2\to FeB{r}_2$
$3FeB{r}_2+B{r}_2\to F{e}_{3}B{r}_8$
$F{e}_{3}B{r}_8+4N{a}_{2}C{O}_3\to 8NaBr+4C{O}_2+F{e}_3{O}_4$

When $8$ moles of $NaBr$ are formed, $4$ moles of carbon dioxide are formed.
The number of moles of carbon dioxide formed is $\frac{2.06\times {10}^3}{102.9}\times \frac{4}{8}=10$.