Let $(G,\ast)$ be a Cyclic group of order $ ‘n\text{’}$: The number of generators of
$G= “\phi(n)\text{”}$.
Euler's function counts the positive integers up to a given integer $n$ that are relatively prime (co- prime) to $n.$
$\textbf{Co-prime:}$ It can be defined more formally as the number of integers $k$ in the range $1\leq k\leq n$ for which the greatest common divisor $gcd(n,k)$is equal to $1.$
If $n=pq(p\: \text{and}\: q\: \text{are distinct prime number})$ then $\phi(n) = \phi(p)\times \phi(q)$.
where $\phi(p) = p-1$ and $\phi(q) = q-1$.
$\textbf{General Formula is :}$ $\phi(p^{n}) = p^{n}-p^{n-1}$, where $p$ is a prime number.
Now $2020 = 2^{2}\times 5^{1} \times 101^{1} $
$\phi(2020) = \phi(2^{2}\times 5^{1} \times 101^{1})$
$\implies \phi(2^{2}) \times \phi(5^{1}) \times \phi(101^{1})$
$\implies (2^{2}-2^{1}) \times (4) \times(100)$
$\implies 2 \times 4 \times 100 = 800$ generators.
So, the correct answer is $800$.
