When you have to find the factors of a number, start by finding the 'prime box' of that number (how many of each prime factor does the number have).
210=70x3=7x10x3=7x2x5x3
So 210 has one of each of the following: 2,3,5,7.
Now, we need to figure out how many combinations we can make of 3,5,7 (we're not allowed to use the 2 because we want only odd factors).
Use combinatorics to figure out the number of combinations:
You have two choices for the number of 3s you use (none or one)
You have two choices for the number of 5s you use (none or one)
You have two choices for the number of 7s you use (none or one)
In total that's 2x2x2=8 combinations.
Don't forget to subtract 1 because the question specified odd factors GREATER than 1, and the combination where you use (none,none,none) for (3,5,7) will give you the number 1 (which of course is a factor of every integer).
It might be easier to just find all of the combinations using brute force, but you run the risk of missing one (and at a higher level of difficulty they would give you a number much bigger than 210).
The answer is E.

It may be just as easy to list all the factors of 210 and count up the odd ones! Avi's method of getting to the prime box ensures that you don't miss any.

1 210
2 105
3 70
5 42
6 35
7 30
10 21
14 15