A tension of 22N is applied to a copper wire of cross sectional area 0.02cm², youngs modulus of copper is $1.1\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2$ and poisson's ratio 0.32. The decrease in cross sectional area will be 
 

We know that,

Young's modulus $\mathrm{Y}=\frac{\mathrm{F}l}{\mathrm{A}∆l}$

Poisson's ratio $\mathrm{\sigma }=\frac{{\displaystyle \frac{∆\mathrm{r}}{\mathrm{r}}}}{{\displaystyle \frac{∆l}{l}}}$

from the above equation,

$\frac{∆\mathrm{r}}{\mathrm{r}}=\frac{\mathrm{F\sigma }}{\mathrm{AY}}=\frac{22\times 0.32}{0.02\times {10}^{-4}\times 1.1\times {10}^{11}}=32\times {10}^{-6}$

$\mathrm{Cross} \mathrm{sectional} \mathrm{area} \mathrm{A}={\mathrm{\pi r}}^2$

$\Rightarrow \mathrm{\Delta A}=2\mathrm{\pi r\Delta r}$

$\Rightarrow \frac{∆\mathrm{A}}{\mathrm{A}}=\frac{2∆\mathrm{r}}{\mathrm{r}}=64\times {10}^{-6}$

$\Rightarrow ∆\mathrm{A}=64\times {10}^{-6}\times 0.02=1.28\times {10}^{-6}{\mathrm{cm}}^2$

Hence the correct is $1.28\times {10}^{-6}{\mathrm{cm}}^2.$