A natural number's square consists of 202 digits. The first 100 digits are 1s, followed by 101 digits of 2s. Determine the last digit and the number.

Let's solve this problem step by step.

1. **Understanding the Problem:**
   - The square of a natural number has 202 digits.
   - The first 100 digits are 1s.
   - The next 101 digits are 2s.

2. **Formulating the Number:**
   - The number can be represented as \( N^2 \), where \( N \) is the natural number we need to find.
   - The number \( N^2 \) looks like \( 111\ldots111222\ldots222 \) (100 ones followed by 101 twos).

3. **Finding the Last Digit:**
   - To find the last digit of \( N^2 \), we need to consider the last digit of \( N \).
   - The last digit of \( N^2 \) is determined by the last digit of \( N \).

4. **Possible Last Digits of \( N \):**
   - The last digit of \( N \) can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.
   - We need to check which of these, when squared, end in 2.

5. **Checking Each Case:**
   - \( 0^2 = 0 \)
   - \( 1^2 = 1 \)
   - \( 2^2 = 4 \)
   - \( 3^2 = 9 \)
   - \( 4^2 = 16 \) (last digit is 6)
   - \( 5^2 = 25 \) (last digit is 5)
   - \( 6^2 = 36 \) (last digit is 6)
   - \( 7^2 = 49 \) (last digit is 9)
   - \( 8^2 = 64 \) (last digit is 4)
   - \( 9^2 = 81 \) (last digit is 1)

   None of these squares end in 2. Therefore, the last digit of \( N \) must be such that when squared, it ends in 2. This is not possible with the digits 0 through 9.

6. **Conclusion:**
   - Since no single digit squared ends in 2, the problem as stated has no solution in the realm of natural numbers.

Therefore, the problem as stated does not have a valid solution. If there is additional context or a different interpretation, please provide it.
Let the sought number be denoted by $S$, and the number consisting of 100 ones by $E$. According to the problem, in $S^{2}$, after $E$ comes twice $E$, then a two and an unknown digit, let's say $x$. Generally, if the $k$-digit number $B$ is written after the number $A$, in the resulting $\overline{A B}$ number, the place value of each digit of $A$ increases by a factor of $10^{k}$, $\overline{A B}=10^{k} A+B$. Based on this, $S^{2}$ can be derived from $E$ as follows:

$$
S^{2}=10\left(10\left(10^{100} E+2 E\right)+2\right)+x
$$

The number $E$ consisting of 100 ones is nine times a number consisting of 100 nines, and by increasing this by one, we get the number $D=10^{100}$, which consists of one one and 100 zeros: $9 E+1=D$, thus

$$
\begin{aligned}
& S^{2}=100(D+2) E+20+x=100(D+2) \frac{D-1}{9}+20+x \\
& 9 S^{2}=100 D^{2}+100 D-20+9 x=(10 D+5)^{2}+9(x-5)
\end{aligned}
$$

This is a perfect square if the value of the second term is 0, that is, $x=5$, in which case $S=(10 D+5) / 3=\left(10^{101}+5\right) / 3$. Since the sum of the digits of the number $C=10^{101}+5$ is 6, it is indeed divisible by 3, and the value of $S$ is an integer. No other value can be considered for $x$, since the squares adjacent to $C^{2}$ differ from $C^{2}$ by a much larger amount than $9(x-5)$ can be for a single-digit $x$. Therefore, the sought number is $\left(10^{101}+5\right) / 3$, and the last digit of its square is 5.