$2025$, i.e. $3^{2025}\mid T$.

$$T=\underbrace{333\cdots 3}_{3^{2024}}=\frac{1}{3}\left(10^{3^{2024}}-1^{3^{2024}}\right)$$

Applying the LTE (# 4802) to the inside term we find $V_3(10-1)+V_3(3^{2024})=2026$. Hence the final answer is $2025$.