Solution

The resistance of a wire is given by $$R=\rho \dfrac{l}{A}$$
where $$\rho$$ is the resistivity of the material
$$l$$ is the length of the wire
$$A$$ is the cross sectional area of the wire
Since stretching of a wire does not change the volume of the wire,
$$V=lA=constant$$
$$\implies l=\dfrac{V}{A}$$
When radius is decreased by $$0.1$$%, the area decreases to $$0.999^2A$$.
Thus the resistance becomes $$R'=\rho \dfrac{V/A'}{A'}=\rho \dfrac{V}{A'^2}$$
$$=\rho \dfrac{V}{0.999^4A^2}$$
$$\implies \dfrac{R'}{R}=\dfrac{1}{0.999^4}$$
Thus percentage increase in resistance=$$\dfrac{R'-R}{R}\times 100$$ %
$$=0.4$$%