2. (7th "Hope Cup" Senior High School Competition) Given the function $f(n)=k$, where $k$ is the $n$-th digit after the decimal point of the repeating decimal $0 . \dot{9} 1827364 \dot{5}$, then the value of $\underbrace{f\{f \cdots f[f}_{1996 \uparrow f}(1)]\}$ is
A. 9
B. 7
C. 3
D. 4
2. D Specifically calculate the first few values of $f$: $f(1)=9$,
$$
\begin{array}{l}
f(f(1))=f(9)=5, f[f(f(1))]=f(5)=7 . \\
f_{1}(1)=f(7)=6, f_{5}(1)=f(6)=3 . \\
f_{1}(1)=f(3)=8 \cdot f_{7}(1)=f(8)=4 . \\
f_{n}(1)=f(4)=2, f_{7}(1)=f(2)=1 .
\end{array}
$$

After 9 applications of $f$, a cycle appears.
Therefore, $f_{1006}(1)=f_{2=1} \times 1,(1)=f_{3}(1)=4$. Hence, the answer is $\mathrm{D}$.