Answer
$1.38\;\rm mm$
Work Step by Step
To find the by how many mm did the G string stretch, we need to use Young’s modulus where the change in length of the wire is given by
$$\Delta L=\dfrac{LF}{Y_{\rm steel}A}\tag 1$$
where we are given the original length of the wire, the diameter of it, and we need to find the force exerted on it which is the tension force applied.
We know that the tesnion force is given by
$$v=\sqrt{\dfrac{T_s}{\mu}}$$
Hence,
$$T_s=\mu v^2$$
Plugging into (1);
$$\Delta L=\dfrac{\mu v^2L}{Y_{\rm steel}\pi r^2} $$
Plugging the known;
$$\Delta L=\dfrac{(1.3\times 10^{-3}) (250)^2(0.75)}{(20\times 10^{10})\pi \left(\frac{0.46}{2}\times 10^{-2}\right)^2}=\bf 1.38\times 10^{-3}\;\rm m $$
$$\Delta L=\color{red}{\bf1.38}\;\rm mm$$
