Mosquito in the Room (Puzzle)

Not really an answer, but it can serve as a starting point.

As noted in the comments (Hagen von Eitzen and ccorn) -and included in the edited question- a swipe volume $V > V_0= x \times y \times d$ (where $x,y$ are the smallest sides of the box and $d$ is the maximum distance travelled by the mosquito; here $x \times y \times d = 3 m \times 10 m \times 1 m=30 m^3$) is enough. One starts swiping a slice adjacent to the smallest face of the box.

In general: assume that before swipe $n$ we knew for sure that the mosquito was not in the excluded volume $C_{n}$, and assume that we swipe (without success) the region $S_n$. Now, we know that the mosquito is not in the region $S_n \cup C_n$; but in the next iteration it can be in any place a distance $d$, hence all we know is that $C_{n+1} = \psi_d(S_n \cup C_n)$, where $\psi_d(X)$ is the erosion morphological operator (it substracts from region $X$ all points that are at distance less than $d$ from its complement $\bar X$).

The goal is to construct an increasing sequence $C_n$ that eventually fills the box. That's possible with the above construction.

Now, suppose that $V<V_0$. Then, it's clear that the above approach (start with a slice on the smaller face and try to make it thicker) does not work: the erosion removes all the excluded volume and we cannot make progress. What remains (what seems difficult) is to show that there is no better strategy, in the long run. For example, one could try starting with a corner (tethraedron) instead of a slice, so that the first erosion leaves us with some non-empty excluded volume, but soon (at least, when we've covered half of the box) we loose that advantage and we cannot progress.

Specifically, it would be enough to prove this (seemingly true) conjecture: Let $A$ be any region inside the box, with volume greater than half of the box (and less than the full box). Then, the "eroded volume" (points with distance less than $d$ from $\bar A$) is at least that of the corresponding slice parallel to the smaller face: $V(A) - V(\psi_d(A)) \ge x\times y\times d $