Answer
The ACL must be stretched by $3.2~mm$ to tear it.
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
In this situation, we can let $\frac{F}{A} = 1.9\times 10^8~Pa$
We can find the change in length:
$\Delta L = (\frac{F}{A})~(\frac{L}{Y})$
$\Delta L = (1.9\times 10^8~Pa)~(\frac{1.0\times 10^{-2}~m}{6.0\times 10^8~Pa})$
$\Delta L = 3.2\times 10^{-3}~m = 3.2~mm$
The ACL must be stretched by $3.2~mm$ to tear it.
