$$
f_{u:-1}(k)=f_{1}\left(f_{u}(k)\right) .
$$

Find $f_{1991}\left(2^{1000}\right)$.

Let $k$ be much smaller, and we can estimate the value range of $f_{1}(k)$ as follows:

Suppose the positive integer $a$ has $m$ digits, then by changing all the digits of $a$ to 9, we have
$$
f_{1}(a) \leqslant 9^{2} m^{2}=81 m^{2}.
$$

If $a \leqslant b$, then $m$ is no greater than the integer part of $\lg b$ plus 1, thus
$$
f_{1}(a) \leqslant 81(1 + \lg b)^{2} < \left(4 \log _{2} 16 b\right)^{2}.
$$

Using the above inequality, we have
$$
\begin{array}{l}
f_{1}\left(2^{1990}\right) < 2^{4} 1994^{2} < 2^{20}, \\
f_{2}\left(2^{1900}\right) < (4 \cdot 30)^{2} = 14400.
\end{array}
$$

Therefore, the sum of the digits of $f_{2}\left(2^{1990}\right)$ is no more than $4 \cdot 9$, so
$$
\begin{array}{l}
f_{3}\left(2^{1990}\right) < 36^{2} = 1296, \\
f_{4}\left(2^{1800}\right) < (9 + 9 + 9)^{2} = 729, \\
f_{5}\left(2^{1000}\right) < (6 + 9 + 9)^{2} = 24^{2}.
\end{array}
$$

On the other hand, we have
$$
f_{1}(k) = k^{2} (\bmod 9).
$$

Since $12^{8} \equiv 1 (\bmod 9)$, we get
$$
\begin{aligned}
f_{1}\left(2^{1900}\right) & \equiv \left(2^{1990}\right)^{2} \equiv \left(2^{4}\right)^{2} \\
& \equiv 4 (\bmod 9), \\
f_{2}\left(2^{1900}\right) & \equiv -2 (\bmod 9), \\
f_{3}\left(2^{1900}\right) & \equiv 4 (\bmod 9).
\end{aligned}
$$

By analogy, generally, we have
$$
f_{0}\left(2^{1900}\right) \equiv \left\{\begin{array}{l}
4 (\bmod 9), 2 \mid n_{3} \\
-2 (\bmod 9), 2 \mid n_{0}.
\end{array}\right.
$$

Thus, since $f_{5}\left(2^{1580}\right) < 24^{2}$, $f_{5}\left(2^{1900}\right) \equiv 4 (\bmod 9)$, and noting that $f_{n}$ is a perfect square, we can deduce
$$
f_{5}\left(2^{1980}\right) \in \{4, 49, 121, 256, 400\}.
$$

For these 5 numbers, we find the square of the sum of their digits, yielding $f_{8}\left(2^{1900}\right) \in \{16, 169\}$.
Continuing to find the square of the sum of the digits, we get
$$
\begin{array}{l}
f_{7}\left(2^{1900}\right) \in \{49, 256\}; \\
f_{8}\left(2^{1090}\right) = 169.
\end{array}
$$

Therefore, when $n \geqslant 8$, we have
$$
f_{n}\left(2^{1000}\right) = \left\{\begin{array}{l}
169, 2 \mid n_{3} \\
256, 2 \nmid n_{0}.
\end{array}\right.
$$

In particular, we have $f_{1001}\left(2^{1000}\right) = 256$.
Solution 2: To quickly obtain the desired result for $f_{0+1}\left(2^{1990}\right)$, we simultaneously consider the remainder of $f_{0}\left(2^{1800}\right)$ modulo 9.

Since $2^{1000} < 8^{700} < 10^{700}$, $2^{1200}$ has at most 700 digits. Thus,
$$
\begin{array}{l}
f_{1}\left(2^{1990}\right) < (9 \cdot 700)^{2} < 4 \cdot 10^{7}, \\
f_{2}\left(2^{1990}\right) < (3 + 9 \cdot 7)^{2} < 4900, \\
f_{3}\left(2^{1990}\right) < (3 + 9 \cdot 3)^{2} = 30^{2}.
\end{array}
$$

Since $f_{3}\left(2^{1900}\right)$ is a perfect square, let $f_{s}\left(2^{1900}\right) = a^{2}$, then $a \equiv f_{2}\left(2^{1090}\right) \equiv 7 (\bmod 9)$. Therefore,
$$
\begin{aligned}
f_{3}\left(2^{1900}\right) & \in \{7^{2}, 16^{2}, 25^{2}\} \\
& = \{49, 256, 625\}.
\end{aligned}
$$

Noting that the sum of the digits of 49, 256, and 625 is 13, we have
$$
f_{4}\left(2^{1900}\right) = 13^{2} = 169.
$$

Continuing, we get
$$
\begin{array}{l}
f_{5}\left(2^{1990}\right) = 16^{2} = 256, \\
f_{0}\left(2^{1990}\right) = 13^{2} = 169.
\end{array}
$$

And so on. Finally, we have
$$
f_{1801}\left(2^{1800}\right) = 256.
$$