Note that $1001 = 11 \cdot 91$. If $11 \mid a$ then also $11 \mid a^{2020} + 10a^{1010} + 1001$. Otherwise, consider $a^5$ modulo $11$. A case study shows that if $a$ is congruent to $1$, $3$, $4$, $5$, or $9$, then $a^5$ is congruent to $1$, and in all other cases, $a^5$ is congruent to $-1$. Hence $a^{10} \equiv 1 \pmod{11}$ whenever $11 \nmid a$. This implies that $a^{1010} \equiv 1 \pmod{11}$ and $a^{2020} \equiv 1 \pmod{11}$ because $a^{1010} = (a^{10})^{101}$ $\pmod{11}$ and $a^{2020} = (a^{10})^{202}$. Consequently, $10a^{1010} \equiv 10 \pmod{11}$, implying that $a^{2020} + 10a^{1010} + 1001 \equiv 0 \pmod{11}$. Thus $a^{2020} + 10a^{1010} + 1001$ cannot be prime since obviously $a^{2020} + 10a^{1010} + 1001 \ge 1001 > 11$.