Solution:

Answer $6$

Suppose one endpoint of a segment length $3$ is at $A$. Evidently the other end could be at the edge midpoints $B$, $C$, $D$. It could also be on the circular arc connecting $B$ and $C$ (with center $O$ and radius $\sqrt{5}$). Similarly, it could be on arcs connecting $C$ and $D$, or $B$ and $D$. We claim that if $X$ is a point of one of these arcs other than its endpoints, then the only possible segment length $3$ with an endpoint at $X$ (and the other endpoint on the surface of the cube) is $AX$. Without loss of generality we can consider $X$ to be on the arc $BC$. Take axes with origin $O$, so that $A$ is $(0,0,2)$. Suppose $X$ is $(a,b,0)$ and that the other endpoint of the segment is $Y$ $(x,y,z)$. Then

$$XY^2 = (x - a)^2 + (y - b)^2 + z^2 = a^2 + b^2 + z^2 - x(2a - x) - y(2b - y) = 5 + z^2 - x(2a - x) - y(2b - y)$$

But $a$, $b > 1$ since $X$ is not an endpoint of the arc, so $(2a- x)$ and $(2b- y)$ are both positive. Hence $- x(2a- x) - y(2b- y) \leq 0$ with equality iff $x = y = 0$. Similarly, $z^2 \leq 4$ with equality iff $z = 2$. Hence $XY^2 \leq 9$ with equality iff $Y = A$, which proves the claim.

Thus if the next link of the polygonal line goes from $A$ to anywhere except $B$, $C$, $D$, then it has to go back to $A$. So a minimal line must go to $B$, $C$, or $D$.

Now from $D$ the line can only go to $A$ or $O$. For if it goes to $Z$ $(x,y,z)$, then we have
$$DZ^2 = (x - 2)^2 + (y - 2)^2 + (z - 1)^2 \leq 2^2 + 2^2 + 1^2 = 3^2$$
with equality iff $x = 0$, $y = 0$ and $z = 0$ or $2$.

So let us take $A$ as the starting point of the polygonal line. Without loss of generality the first segment is $AD$. Then the second segment must be $DO$ (for a minimal line). Thus the best we can do with $2$ segments is to move along an edge. It takes three such moves to get to the opposite corner, and hence at least $6$ segments. But it is obvious that it can be done with $6$ segments.