Try without finding the sum...

$\displaystyle\sum_{k=1}^{99} k^2 \equiv 11 \displaystyle\sum_{k=1}^9 k^2\pmod{9}$

because

$(9m+1)^2+(9m+2)^2+\cdots (9m+9)^2 \equiv 1^2+2^2+\cdots+9^2 \pmod{9}$ for any $m \in \mathbb Z$ $\equiv 4 \displaystyle\sum_{k=1}^4 k^2\pmod{9}$ because $k^2 \equiv (9-k)^2\pmod{9})$ $\equiv 3\pmod{9}.$ The remainder is 3.

$1^2+2^2+3^2+...+99^2=\frac{99\times100\times199}6=33\times50\times199$

$33\times50\times199\equiv6\times5\times1\equiv3\pmod9$