Of course both lengths are contracted from the other point of view.
The question was a bit poorly worded, as in relativistic settings "at the same time" should never be said without saying for which observer it is at the same time. I assumed that it is at the same time for the alien pressing the button.
So the alien presses the button at (ct,x) = (0,0). The center of the spaceship also has the position (0,0). We know, that the front of the spaceship is 5m from the center in the system of the spaceship, so the button press event is at (ct, 5m). In the system of the alien, both probes are released at the time 0, so it's at (0, x). Using the Lorentz transformation, we get x=2.5m, and ct=-4.33m. The same considerations can be applied to the back, and there x=-2.5m and ct=4.33m
The answer, how far the pods are apart for the alien is thus 5m.
As I said, the length contraction goes both ways. So the distance between the two pods is only 2.5m for the pilot of the spaceship. This seems to be a contradiction to the fact that the spaceship is 10m long. But it's not, because the probes are not released at the same time in the spaceship system. The pilot sees the front probe released first, and then 8.66m / c later the back probe is released. In this time, the spaceship traves 7.5m. As the probes are 10m apart on the spaceship, that gives exactly the 2.5m distance between the released probes.
(And then there is the question, how much of the probes will be left for the second alien to measure, after they hit the ground at 0.866c. I guess he will have to measure the distance between the craters.)