Given,

The area of cross-section of copper wire is $A=0.01 {\mathrm{cm}}^2$

Tension in the wire is $T=22 \mathrm{N}$

Young's modulus of copper wire is $Y=1.1\times {10}^{11} \mathrm{N} {\mathrm{m}}^{-2}$

Poisson ratio is $\sigma =0.32$

Using the formula of Young's modulus of wire, 

$Y=\frac{F}{A}\times \frac{l}{dl}$

The longitudinal strain is given by,

$\Rightarrow \frac{dl}{l}=\frac{T}{AY}$

$\Rightarrow \frac{dl}{l}=\frac{22}{{10}^{-6}\times 1.1\times {10}^{11}}=20\times {10}^{-5}$

The Poisson ratio of wire is given by,

$\sigma =\frac{{\displaystyle \frac{dr}{r}}}{{\displaystyle \frac{dl}{l}}}$

The lateral strain in the wire is given by

$\Rightarrow \frac{dr}{r}=\sigma \times \frac{dl}{l}=0.32\times 20\times {10}^{-5}=6.4\times {10}^{-5}$

The percentage change in area is given by

$\Rightarrow \frac{dA}{A}\times 100=\frac{2dr}{r}\times 100=2\times 6.4\times {10}^{-3}$

$\Rightarrow \frac{dA}{A}\times 100=12.8\times {10}^{-3}$