On test day, I'd probably first find the prime factorization of 240 to get: 240 = (2)(2)(2)(2)(3)(5)

---------ASIDE-----------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:

If k is a factor of N, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)
-----BACK TO THE QUESTION!---------------------

Let's check each answer choice...
(A) 6. Since 240 = (2)(2)(2)(2)(3)(5), we can see that 6 is a factor of 240
(B) 9. For 9 to be a factor, we'd need two 3's to be hiding in the prime factorization. There aren't two 3's
(C) 12. Since 240 = (2)(2)(2)(2)(3)(5), we can see that 12 is a factor of 240
(D) 20. Since 240 = (2)(2)(2)(2)(3)(5), we can see that 20 is a factor of 240
(E) 36. For 36 to be a factor, we'd need two 3's and two 2's to be hiding in the prime factorization. There aren't two 3's
(F) 45. For 45 to be a factor, we'd need two 3's and one 5 to be hiding in the prime factorization. There aren't two 3's

Answer: A, C, D