Answer: C

Small Hint:

$\omega = \dfrac{-1 + i\sqrt3}{2}$ is a primitive cube root of unity, so nonzero roots come in triples $\{r, \omega r, \omega^2 r\}$

Big Hint:

With five roots the distinct roots form one triple with multiplicities summing to $5;$ real coefficients force conjugate symmetry

Solution:

Here $\omega = \tfrac{-1 + i\sqrt3}{2}$ is a primitive cube root of unity. Since $0$ is not a root, the set of distinct roots is closed under multiplication by $\omega,$ so it consists of triples $\{r, \omega r, \omega^2 r\}$ equally spaced in argument. Five roots cannot fill two such triples, so there is exactly one triple, with multiplicities $m_1, m_2, m_3 \ge 1$ summing to $5.$

Real coefficients require the root multiset to be closed under conjugation. This is possible only when the triple’s arguments are symmetric about the real axis, which happens for the two configurations $\{0^\circ, 120^\circ, 240^\circ\}$ and $\{60^\circ, 180^\circ, 300^\circ\}.$

The product of the roots must equal $-2020.$ In the first configuration the real root is positive, forcing a positive product, which is impossible. In the second, the real root is negative and the product is $-\rho^5;$ setting $\rho^5 = 2020$ works, and the two conjugate-symmetric multiplicity patterns $(1, 3, 1)$ and $(2, 1, 2)$ each give a valid polynomial. Hence there are $2.$

Thus, the correct answer is C.