Let initial length of the wire be $l$. After stretching new length of the wire $l^{\prime}$ is given as
$$
l^{\prime}=l+10 \% \text { of } l=l+\frac{10}{100} l=\frac{11 l}{10}=1 . l
$$
i.e. length is increased by 1.1 times
$$
\therefore \quad n=1.1
$$
We know that, when length of a wire is increased by $n$ times, then its new resistance is increased by $n^2$ times i.e. $R^{\prime}=n^2 R=(1.1)^2 R \quad[\because$ Here, $n=1.1]$ $=1.21 \mathrm{R}$

Specific resistance (resistivity) of a wire does not depends on its dimensions (length, width etc.) because it is a characteristic property of a materia and depends only nature of material of wire. Hence, specific resistance remains same.