1. **Problem statement:** We want to find the integer $a$ in the fourth term of the binomial expansion of $(1.02)^6$, given the first term is 1 and the fourth term is $(0.000008)a$. 2. **Recall the binomial expansion formula:** $$ (1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k $$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 3. **Identify values:** Here, $n=6$ and $x=0.02$ because $1.02 = 1 + 0.02$. 4. **Find the fourth term:** The terms are indexed from $k=0$, so the fourth term corresponds to $k=3$: $$ T_4 = \binom{6}{3} (0.02)^3 $$ 5. **Calculate the binomial coefficient:** $$ \binom{6}{3} = \frac{6!}{3!3!} = \frac{720}{6 \times 6} = 20 $$ 6. **Calculate $(0.02)^3$:** $$ (0.02)^3 = 0.02 \times 0.02 \times 0.02 = 0.000008 $$ 7. **Calculate the fourth term:** $$ T_4 = 20 \times 0.000008 = 0.00016 $$ 8. **Express the fourth term as $(0.000008)a$:** $$ 0.00016 = (0.000008) a \implies a = \frac{0.00016}{0.000008} = 20 $$ **Final answer:** $$ \boxed{20} $$