The value of $$\frac{625}{4} p_1$$ is .........

Each draw is uniform on $$\{1,2,\dots ,100\}$$ and the three draws are independent.

Step 1 Find $$p_1$$.
Let $$M$$ be the maximum of the three chosen numbers.
It is easier to use the complement: $$P(M \ge 81)=1-P(M\le 80)$$.

For $$M\le 80$$, every draw must fall in $$\{1,2,\dots ,80\}$$.
Probability that one draw is $$\le 80$$ is $$\frac{80}{100}=0.8$$.
Because the draws are with replacement,

$$P(M\le 80)=0.8^3=0.512.$$

Hence
$$p_1 = 1-0.512 = 0.488.$$

Step 2 Multiply by $$\frac{625}{4}$$.
First note $$\frac{625}{4}=156.25$$.

Compute
$$\frac{625}{4}\,p_1 = 156.25 \times 0.488.$$ Using simple multiplication:
$$156.25 \times 0.488 = 156.25 \times \left(0.5-0.012\right) = 78.125 - 1.875 = 76.25.$$

Answer: $$\frac{625}{4}\,p_1 = 76.25,$$ which lies in the required range $$76.10-76.40$$.