1. Interpret the experiment

Three numbers are chosen with replacement from $S=\{1,2,3,\dots,100\}.$ So each draw is independent, and each number from $1$ to $100$ is equally likely.

We need $p_1 =$ probability that the maximum of the three chosen numbers is at least $81$.

Then we must compute $\frac{625}{4}p_1.$

2. Find $p_1$ using the complement

The event $\max\{X_1,X_2,X_3\} \ge 81$ is easier to handle by using its complement: $\max\{X_1,X_2,X_3\} < 81.$

If the maximum is less than $81$, then all three numbers must lie in $\{1,2,3,\dots,80\}.$

For one draw, $P(\text{number} \le 80)=\frac{80}{100}=\frac45.$

Since the draws are independent, $P(\max<81)=\left(\frac45\right)^3=\frac{64}{125}.$

Therefore, $p_1=1-\frac{64}{125}=\frac{61}{125}.$

3. Compute $\frac{625}{4}p_1$

$\frac{625}{4}p_1=\frac{625}{4}\cdot \frac{61}{125}.$

Simplify: $\frac{625}{125}=5,$ so $\frac{625}{4}\cdot \frac{61}{125}=\frac{5\cdot 61}{4}=\frac{305}{4}=76.25.$

4. About $p_2$

The quantity $p_2$ is defined in the question, but it is not needed for the asked value.

5. Final answer

$\frac{625}{4}p_1=76.25$