Solution:

The answer is (D). Let $L$ be the point on the floor where the thief "lands", $D$ the point where the diamond is located; let $D_{E}, D_{W}, D_{N}, D_{S}$ be the reflections of $D$ with respect to the east, west, north, south walls $(E, W, N, S$ in the figure). Let us imagine that the thief touches the wire (in order to cut it) at a point $P$ on wall $E$, and then goes to fetch the diamond at $D$: the path he takes has length (at least) $L P+P D$, which by symmetry is equal to $L P+P D_{E}$; hence, assuming that the thief cuts the wire at a point on wall $E$, the shortest possible path, as a consequence of the triangle inequality, is the one in which $P$ is aligned with $L$ and $D_{E}$ (this point is denoted by $P_{E}$ in the figure), and it has length $\sqrt{4^{2}+1}=\sqrt{17}$ meters.

Repeating the same reasoning for the other three walls (and noting, in order to save calculations, that $L D_{E}=L D_{W}, L D_{S}=L D_{N}$) we reduce to comparing the lengths of $L D_{E}$ and $L D_{N}$; it follows that the minimum distance to be traveled is $\sqrt{2^{2}+3^{2}}=\sqrt{13}$ meters.