Answer
$$
E=-5.5 \mathrm{J}
$$
Work Step by Step
At $x=2.0 \mathrm{m},$ we estimate the potential energy to be
$$
U(x=2.0 \mathrm{m}) \approx U(x=1.0 \mathrm{m})+(-4.9 \mathrm{J} / \mathrm{m})(1.0 \mathrm{m})=-7.7 \mathrm{J}
$$
Thus, the total mechanical energy is
$$
E=K+U=\frac{1}{2} m v^{2}+U=\frac{1}{2}(2.0 \mathrm{kg})(-1.5 \mathrm{m} / \mathrm{s})^{2}+(-7.7 \mathrm{J})=-5.5 \mathrm{J}
$$
Again, there is some uncertainty in reading the graph which makes the last digit not very significant.
At that level $(-5.5 \mathrm{J})$ on the graph, we find two points where the potential energy curve has that value $-$ at $x \approx 1.5 \mathrm{m}$ and $x \approx 13.5 \mathrm{m} .$
Therefore, the particle remains in the region $1.5
