Some thoughts ..
(a) Most of the question is pointless mathematics to confuse you, and has very little to do with how you would calculate mass and balance.
The arithmetic is simple and precisely how you do calculate mass and balance sums (although you appear to know that from the remainder of your commentary) ... the question simply tries to make a fairly boring moment sum a little bit more interesting .. even instructors and examiners get a tad tired of the same old same old all the time ...
(b) presuming that the isolated wheel loads are uniform, the nose and main loads can be assumed to act through the respective centroids (as has been observed in several posts)
The solution is very simple if some of the very basics are kept in mind ... BOAC's exhortation to draw a piccy, invariably, is good advice ... if I may be so bold as to simplify his comments/solution a tad further ...
(a) note that the position of the calculation datum has no effect on the significance of the answer (although the numbers will relate to whatever datum is chosen)
(b) we are interested in finding out the distance of the CG ahead of the mains ... so, would it not make sense to put the datum at the main centroid load ? This also has the advantage of making the moment sum only involve the nose assembly moment (as the main is at the datum .. ie zero moment regardless of load).
Hence ..
CG = total moment/total mass
= (1450*10) / 25450
= 0.57 m
= 57 cm in front of the mains.
either you keep in mind that the CG is in front of the mains, or you can be very clever and use a minus arm for the nose assembly load .. or however you like to keep track of the sense of the moments ...
[no more hectalitres please .. bring back the slug and poundal, I say !!]