Numbers!

I have a objective approach. Sum of that number is 42, it is in form of 3 \times n so it's square will be multiple of 9. If it multiple of 9, then its sum of digit is also multiple of 9, there is a single option, which is multiple people 9,that is 108.

Ez write 654000003987000^2 as ((654000000000+3987) 1000)^2. Because the 1000 doesn't matter because those all get turned into zeroes, and the 10^9 margin gap between 654000000000 is too large, and those zeroes also don't matter, the answer is just the sum of the digits of 654^2 , 3987^2 and 2 654 3987. Because this is probably a Target round, you can probably do it on your calculator. But if not, then write 3987 as 4000-13 and solve from then. But anyway, 3987^2= 15896169, and 1+5+8+9+6+1+6+9=45, and 654^2= 427716, and 4+2+7+7+1+6=27, and 2 654*3987= 5214996, and 5+2+1+4+9+9+6= 36, 45+27+36=108, so 108 is the answer.

There might be another methodology to doing the problem, but I do not see any Number Theory approaches to sum of digits of an power.