Solution:

The answer is 325. Let $ABCD$ be the rectangle that makes up the chocolate wrapper; suppose that $AB$ has length $360~\mathrm{mm}$ and that Anacleto makes point $A$ coincide with the midpoint of side $BC$, which we will call $M$. To do this he must fold along the perpendicular bisector of segment $AM$, which intersects sides $AB$ and $CD$ respectively at points $S$ and $T$; what the problem asks for is the length of segment $ST$. Let $K$ be the intersection of $ST$ and $AM$, and $H$ the projection of $T$ onto $AB$. Triangle $THS$ is similar to triangle $ABM$: both are right triangles, and angle $TSA$ is congruent to angle $AMB$ (this is because it is supplementary to angle $TSB$, which in turn, in order for the sum of the interior angles of $SBMK$ to be $360^\circ$, since $SBM$ and $MKS$ are right angles, must be supplementary to $AMB$). We thus have the similarity $TH : TS = AB : AM$. The length of $AB$ is $360~\mathrm{mm}$, that of $TH$ (equal to that of $AD$) is $300~\mathrm{mm}$, and that of $AM$ turns out, by the Pythagorean theorem, to be (in millimeters) $\sqrt{AB^2 + AM^2} = \sqrt{360^2 + 150^2} = 30 \sqrt{144 + 25} = 30 \cdot 13$.

We thus obtain the length in millimeters of $TS$: $TS = \frac{AM \cdot TH}{AB} = \frac{30 \cdot 13 \cdot 300}{360} = 325$.