Answer
face 2
Work Step by Step
We know that
$F=Y\frac{\Delta L}{L}A$
This equation shows that $\Delta L\propto \frac{1}{A}$. Thus, the change in length will be minimum, if the brick is placed on the face having the greatest area, that is face 2 whose area is $2cm\times 3cm=6cm^2$
