For any positive integer $k$, let $f_{1}(k)$ be the square of the sum of the digits of $k$ when written in decimal notation, and for $n>1$, let $f_{n}(k)=f_{1}\left(f_{n-1}(k)\right)$. What is $f_{1992}\left(2^{1991}\right)$?
We can answer the question based on two aspects: on the one hand, we determine the number of digits, and on the other hand, the remainder of the number in question when divided by 9.

Let's denote the number $2^{1991}$ as $K$.

$$
K=8^{\frac{1991}{3}}<10^{664}
$$

Thus, the number $K$ has at most 664 digits, and the sum of its digits is no more than $664 \cdot 9$, which is less than 6000. Therefore, $f_{1}(K)<36 \cdot 10^{6}$. Up to this limit, the sum of the digits of the number 29999999 is the largest, 65. Thus,

$$
f_{2}(K) \leq 4225
$$

Similarly,

$$
f_{3}(K) \leq(3+3 \cdot 9)^{2}=900
$$

For a number $k$ with at most three digits, $f_{1}(k) \leq(3 \cdot 9)^{2}=729$, so it is also at most three digits. Therefore, if $n \geq 3$, then $f_{n}(K)$ is at most three digits.

It is known that a number divided by 9 gives the same remainder as the sum of its digits; furthermore, the remainder of the product of two numbers when divided by 9 is the same as the product of their remainders, since $(9 r+s)(9 t+u)=$ $9(9 r t+r u+s t)+s u$. Based on these,

$$
K=2^{6 \cdot 331+5}=64^{331} \cdot 32=(9 \cdot 7+1)^{331}(9 \cdot 3+5)
$$

has a remainder of 5, and thus the remainder of the sum of its digits is also 5. The remainder of $f_{1}(K)$ is therefore the same as that of 25, which is 7; the remainder of $f_{2}(K)$ is the same as that of 49, which is 4; the remainder of $f_{3}(K)$ is the same as that of 16, which is 7. This is also the remainder of the sum of its digits, and the number is at most three digits, so the sum of the digits can only be 7, 16, and 25. The possible values of $f_{4}(K)$ are 49, 256, and 625. The value of $f_{5}(K)$ is the same in all three cases: 169. Continuing the calculation,

$$
f_{6}(K)=256, \quad f_{7}(K)=169, \quad f_{8}(K)=256
$$

It is clear that from here, for even indices we get 256, and for odd indices, 169. The answer to the question is therefore

$$
f_{1992}\left(2^{1991}\right)=256
$$

Remarks. 1. Of course, for other $a$, $c$, and $n$, $f_{n}\left(a^{c}\right)$ can also be determined similarly, and it is easy to see that for sufficiently large $n$, only 1 and 81 can occur besides 169 and 256.

2. The sum of the digits of $2^{1991}$ can be determined using a computer, which is 2669. From this, $f_{1}(K)=7123561, f_{2}(K)=625$, $f_{3}(K)=169, f_{4}(K)=256$, and from here, the last two values repeat periodically.