3. Let $[x]$ denote the greatest integer not exceeding $x$. Then the last two digits of $\left[\frac{10^{2006}}{10^{99}-3}\right]$ are $\qquad$ .
3.23.

From $10^{2000}=10^{59 \times 34}-3^{34}+3^{34}$, we know
$$
\left[\frac{10^{2006}}{10^{99}-3}\right]=\frac{10^{99 \times 34}-3^{34}}{10^{59}-3}=\sum_{k=0}^{33} 10^{59(33-k)} \times 3^{k} \text {. }
$$

Therefore, the last two digits we are looking for are the remainder of $3^{33}$ modulo 100.
Also, $3^{32}=(10-1)^{16}=-\mathrm{C}_{16}^{15} \times 10+1(\bmod 100)$ $=41(\bmod 100)$,
so $3 \times 3^{32}=3 \times 41(\bmod 100) \equiv 23(\bmod 100)$.
Thus, the last two digits we are looking for are 23.