Answer
The string should be stretched by $5.0~mm$
Work Step by Step
$Y = \frac{F/A}{\Delta L/L}$
$Y$ is Young's modulus
$F$ is the force
$A$ is the cross-sectional area
$\Delta L$ is the change in length
$L$ is the original length
We can find the required change in length:
$\Delta L = \frac{F~L}{Y~A}$
$\Delta L = \frac{(20~N)(0.50~m)}{(2.0\times 10^9~N/m^2)(1.0\times 10^{-6}~m^2)}$
$\Delta L = 5.0\times 10^{-3}~m = 5.0~mm$
The string should be stretched by $5.0~mm$
