Suppose there are $n-1$ $8$'s between $20$ and $21$, for some integer $n\geq 2$. Then, the expression becomes
\begin{align*}
20\cdot 10^{n+1} + (8\cdot 10^n + 8\cdot 10^{n-1} + \cdots + 8\cdot 10^2) + 21 &= 200\cdot 10^n + 800\left(10^{n-2} + 10^{n-3} + \cdots + 1\right) + 21 \\ &= 200\cdot 10^n + 800\left(\frac{10^{n-1}-1}{9}\right) + 21 \\ &= 200\cdot 10^n + \frac{80\cdot 10^n - 800}{9} + 21 \\ &= \frac{1880\cdot 10^n - 611}{9} \\ &= 47\left(\frac{40\cdot 10^n - 13}{9}\right). \end{align*}
Since $40\cdot 10^n - 13 \equiv 40 - 13 \equiv 27\equiv 0\pmod{9}$, $\frac{40\cdot 10^n-13}{9}$ is an integer, and since $n\geq 2$, the expression is at least $\frac{4000-13}{9} = 443 > 1$, so the expression is a product of two integers greater than $1$ for all $n$, so it is composite, as desired.