The asteroid is a perfect sphere so the perimeter is given by `P = 2 pi r ` where `r ` is the radius of the sphere.
The landing craft is 10km due 'North' of the drilled hole. Where 'North' is isn't relevant here, as all we need to know is that the landing craft and hole are 10km apart on the (curved) surface of the asteroid. That is, they are 10km apart on a particular perimeter of the asteroid (there are infinitely many on a sphere), where a perimeter is the biggest possible circle drawn about the spherical asteroid, a line dividing it exactly in two (into two hemispheres). This perimeter described here goes through the 'North pole' of the asteroid, but it isn't relevant where that is exactly.
We are told the Sun is directly above the drilled hole in question, and the shadow of the antenna on the landing craft on the ground determines that the Sun is 15 degrees off being directly above the landing craft.
Therefore we can deduce that the 10km on the curved surface of the asteroid corresponds to turning 15 degrees about the very centre of the asteroid (its core). There are 360 degrees in a complete circle, so 15 degrees is just 15/360 = 1/24th of the entire perimeter. So the 10km between the drilled hole and the landing craft represents only 1/24th of the entire perimeter of the asteroid, so that the perimeter must be given by
P = 10km x 24 = 240km
Now, since we have from above that `P = 2 pi r ` where `r ` is the radius of the asteroid, we can rearrange this to get
i) The radius of the asteroid as
`r = P/(2 pi) = 240/(2 pi) approx 240/(2 times 22/7) = 60/(11/7) = 420/11 = 38.2km ` (to 3sf)
ii) The Earth's radius on the other hand is 6371km which is
`6371/38.2 = 167 ` times larger (to 3sf) than the asteroid in question.