Solution:

The ones digit of $n^2$ depends only on the ones digit of $n,$ so the list repeats every $10$ terms: $1, 4, 9, 6, 5, 6, 9, 4, 1, 0.$ One block sums to $45.$ Now $2025 = 202 \cdot 10 + 5,$ so we get $202$ full blocks plus the first five terms: $202 \cdot 45 + (1 + 4 + 9 + 6 + 5) = 9090 + 25 = 9115.$ Therefore, the answer is D.