The answer is 546.
Solution: Prime factorization of your number is
N = 20020¹² = 2²⁴ × 5¹² × 7¹² × 11¹² × 13¹².

So, the set of all divisors of N would be
D = {2ᵃ×5ᵇ×7ᶜ×11ᵈ×13ᵉ | a, b, c, d, e ϵ ℕ, a ≤ 24, b, c, d, e ≤ 12}.

And since each assignment of values to a, b, c, d, e gives a unique number, there would be a total of (24+1)×(12+1)×(12+1)×(12+1)×(12+1) = 714025 different divisors of N.

But to get a 7 in the units digit, no divisor should be a multiple of 2 or 5. Removing them, we have the set of divisors as
D' = {7ᵃ×11ᵇ×13ᶜ | a, b, c ϵ ℕ, a, b, c ≤ 12}, which are
(12+1)×(12+1)×(12+1) = 2179 in number. But not all such divisors will have 7 as their units digit. So we'd analyse a bit more.

We need not worry about the factor 11, since it does not affect the units digit. The problem boils down to finding all multiples formed by combinations of 7 and 13, such that the units digit is 7. If we only look only at the units digit, what combinations of 7 and 3 give us 7? Thus, we need to find
S = {(a, b) | u(3ᵃ×7ᵇ) = 7, a, b ϵ ℕ, a, b ≤ 12}, where u(·) means "units digit".

Now, units digit, u(3ᵃ×7ᵇ) = 7 only in the following cases:

Case 1: u(3ᵃ) = 1 and u(7ᵇ) = 7
Case 2: u(3ᵃ) = 7 and u(7ᵇ) = 1
Case 3: u(3ᵃ) = 3 and u(7ᵇ) = 9
Case 4: u(3ᵃ) = 9 and u(7ᵇ) = 3

But, since a, b ≤ 12, we have:

u(3ᵃ) = 1 if and only if a ϵ {0, 4, 8, 12}
u(7ᵇ) = 1 if and only if b ϵ {0, 4, 8, 12}
u(3ᵃ) = 7 if and only if a ϵ {3, 7, 11}
u(7ᵇ) = 7 if and only if b ϵ {1, 5, 9}
u(3ᵃ) = 3 if and only if a ϵ {1, 5, 9}
u(7ᵇ) = 3 if and only if b ϵ {3, 7, 11}
u(3ᵃ) = 9 if and only if a ϵ {2, 6, 10}
u(7ᵇ) = 9 if and only if b ϵ {2, 6, 10}

Thus,

1. For case 1, we have a ϵ {0, 4, 8, 12} and b ϵ {1, 5, 9}; 4 × 3 = 12
   
2. For case 2, we have a ϵ {3, 7, 11} and b ϵ {0, 4, 8, 12}; 3 × 4 = 12
   
3. For case 3, we have a ϵ {1, 5, 9} and b ϵ {2, 6, 10}; 3 × 3 = 9
   
4. For case 4, we have a ϵ {2, 6, 10} and b ϵ {3, 7, 11}; 3 × 3 = 9
   

having 12, 12, 9, and 9 feasible combinations respectively. So, we have a total of 12+12+9+9 = 42 combinations. Now, considering the 13 possible multiples of 11 (including 11⁰, 11¹, ..., 11¹²), we get 42 × 13 = 546 divisors altogether, which have last digit as 7.
Thus, the answer is 546.