6. Let \( N_{k} = 2007^{2007^{\cdots 3007}} \} k \) times. The problem requires the last two digits of \( N_{2007} \).
\[
N_{2007} = 2007^{\mathrm{N}_{20016}} = (2000 + 7) \cdot \mathrm{N}_{20066} = 2000 \times M + 7^{N_{\text{SMom6}}}
\]

where \( M \) is a positive integer. Therefore, the last two digits of \( N_{2007} \) are the same as those of \( 7^{N_{2000}} \). First, let's observe the pattern of the last two digits of \( 7^n \).
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & $\cdots$ \\
\hline The last two digits of $7^n$ & 49 & 43 & 01 & 07 & 49 & 43 & 01 & 07 & $\cdots$ \\
\hline
\end{tabular}
The last two digits of \( 7^n \) repeat every 4 numbers.
\[
\begin{array}{l}
=4 M_{1}-1, \\
=4\left(M_{1}-1\right)+3. \\
\end{array}
\]
( \( M_{1} \) is a positive integer)

Thus, \( 7^{\mathrm{N}_{2 M M 5}} = 7^{K M M_{1}-1 N+3} \) has the same last two digits as \( 7^3 \), which are 43. Therefore, the answer is C.
6. (2007 Zhejiang Province High School Mathematics Competition A Paper) The last two digits of $\left.2007^{20077}\right\}_{20077}$ are ( )
A. 01
B. 07
C. 43
D. 49