$n(A\cup B) = n(A)+n(B)- n(A\cap B)$
$n(A) = 2020$
So, $t_n = t_1 + (n-1)d,$ where $t_1 = 1,n = 2020$ and $d = 2$
$\implies t_n = 1 +2019\times 2 = 4039$
$\implies A=1,3,5,7,9,11,13,15,17,19,21,\dots, 4039$
$n(B) = 2020$
So, $t_n = t_1 + (n-1)d,$ where $t_1 = 5,n = 2020$ and $d = 5$
$\implies t_n = 5 +2019\times 5 = 10100$
$\implies B=5,10,15,20,25,30,\dots ,10100$
$C=A\cap B = 5,15,25,\dots, 4039 $
$5+(n-1)10\leq 4039$
$10n-5\leq 4039$
$10n\leq 4044$
$n\leq 404.4$
$n=404$
$n(C) = n(A\cap B) = 404$
Therefore, $n(A\cup B) = 2020+2020-404 = 3636$
So, the correct answer is $3636$.