#### Answer

The string is stretched by 1.0 cm

#### Work Step by Step

We can use Young's modulus to solve this question as:
$Y = \frac{F/A}{\Delta~L/L} = \frac{F~L}{A~\Delta L}$
For steel, $Y = 200\times 10^9~N/m^2$.
We can find the the distance $\Delta L$ that the string is stretched;
$Y = \frac{F~L}{A~\Delta L}$
$\Delta L = \frac{F~L}{A~Y}$
$\Delta L = \frac{(2000~N)(0.80~m)}{(\pi)(5.0\times 10^{-4}~m)^2~(200\times 10^9~N/m^2)}$
$\Delta L = 0.010~m = 1.0~cm$
The string is stretched by 1.0 cm