6. 15 For any positive integer $k, f_{1}(k)$ represents the square of the sum of the digits of $k$, and
$$
f_{n}(k)=f_{1}\left(f_{n-1}(k)\right), n \geqslant 2 .
$$

Find: $f_{1988}(11)$.
［Solution］From the given conditions, we have
$$
\begin{array}{l}
f_{1}(11)=(1+1)^{2}=4, \\
f_{2}(11)=f_{1}(4)=4^{2}=16 \\
f_{3}(11)=f_{1}(16)=(1+6)^{2}=49, \\
f_{4}(11)=f_{1}(49)=(4+9)^{2}=169, \\
f_{5}(11)=f_{1}(169)=(1+6+9)^{2}=256, \\
f_{6}(11)=f_{1}(256)=(2+5+6)^{2}=169, \\
f_{7}(11)=f_{1}(169)=256 .
\end{array}
$$

From this, we can see that when $n \geqslant 4$, the value of $f_{n}(11)$ alternates regularly, that is,
$$
f_{n}(11)=\left\{\begin{array}{ll}
169, & n \text { is an even number greater than } 3, \\
256, & n \text { is an odd number greater than } 3.
\end{array}\right.
$$

Therefore, $f_{1988}(11)=169$.