Answer
$2.7\ m$, yes.
Work Step by Step
Step 1. Draw a diagram to illustrate the configuration, let $x$ be the width of the boarder in meters.
Step 2. The larger rectangle has a length of $12+2x$ and a width of $8+2x$
Step 3. The boarder area is the difference between the two rectangles, we have $A=(12+2x)(8+2x)-(12)(8)=4x^2+40x\ m^2$
Step 4. Let $A=120$, we have $4x^2+40x=120$ or $x^2+10x-30=0$ which gives $x=\frac{-10\pm\sqrt {100+120}}{2}$. As $x\gt0$, we have $x=\frac{\sqrt {240}-10}{2}=2\sqrt {15}-5\approx2.7\ m$.
Step 5. As $x=2.7\gt2\ m$, the boarder can be done with the existing tiles.
