Solution:

For each $m,$ $\lfloor\sqrt{n}\rfloor = m$ on the $2m + 1$ integers $m^2 \le n \le (m + 1)^2 - 1.$ Since $\sqrt{2025} = 45,$ the terms with $1 \le m \le 44$ contribute $\sum_{m=1}^{44} m(2m + 1),$ and $n = 2025$ tacks on $45.$ That sum is $\sum_{m=1}^{44}(2m^2 + m) = 2 \cdot \tfrac{44 \cdot 45 \cdot 89}{6} + \tfrac{44 \cdot 45}{2} = 58740 + 990 = 59730,$ so the total is $59775.$ Its ones digit is $5.$ Therefore, the answer is D.