If a copper wire is stretched to make it $0.1\%$ longer, the percentage change in its resistance is 

Resistance (R): Resistance is the resistance provided to the passage of electricity.
The ohm is the resistance unit used in SI$\left(\Omega \right)$.
The mathematical definition of resistance is
$R=\frac{\rho l}{A}$
Where $\mathrm{R}=$ resistance, $\mathrm{I}=$ length, $\mathrm{A}=$ area of cross-section, and $\rho =$resistivity

Allow the length of the copper wire to be determined before stretching $\left(I_1\right)=L$
Before-stretching cross-sectional area of copper wire $\left({\mathrm{A}}_1\right)=\mathrm{A}$
The copper wire's resistance prior to stretching $\left(R_1\right)=R$
$R=\rho \frac{L}{A}$
length of the stretched copper wire $\left(I_2\right)=L^{\text{'}}$
Area of a stretched copper wire's cross-section $\left(A_2\right)=A^{\text{'}}$
The resistance of a stretched copper wire $\left(R_2\right)=R^{\text{'}}$
Here,
$L^{\text{'}}=L+0.1\%L=L+\frac{0.1}{100}L=1.001L$
The area of the cross-section will shrink as the length increases. However, the volume is unchanged.

$\therefore \mathrm{V}={\mathrm{V}}^{\text{'}}$
$\Rightarrow \mathrm{AL}={\mathrm{A}}^{\text{'}}{\mathrm{L}}^{\text{'}}$
$\Rightarrow \mathrm{AL}={\mathrm{A}}^{\text{'}}(1.001)\mathrm{L}$
$\Rightarrow A^{\text{'}}=\frac{A}{1.001}$

New resistance will be ${\mathrm{R}}^{\text{'}}$
$R^{\text{'}}=\rho \frac{l^{\text{'}}}{A^{\text{'}}}$
Currently, the resistance's percentage change is
$\frac{\Delta R}{R}=\frac{R^{\text{'}}-R}{R}\times 100$
$\Rightarrow \frac{\Delta R}{R}=\frac{(1.002)R-R}{R}\times 100$

$\Rightarrow \frac{\Delta R}{R}=0.002\times 100=0.2\%$
$\therefore$ The resistance will rise by $0.2\%$.

Hence, the correct answer is option A.