Once you've convinced yourself that cl((12)(34)) = {(a b)(c d) : a, b, c and d any distinct integers from 1 to n}, it's "just" a counting problem.

So fix n. The number of transpositions (a b) in S_n is n(n-1)/2 -- there are n choices for a, n - 1 choices for b,

but (a b) = ( b a) so n(n-1) needs to be divided by 2. Next for the product of 2 transpositions (a b)(c d), there are

n(n-1)/2 choices for the "first" (a b) and (n-2)(n-3)/2 choices for (c d), but (a b)(c d) = (c d)(a b) so we need to divide the product by 2. So there are n(n-1)(n-2)(n-3)/8 elements in the conjugate class. Example: in S_4, there are exactly 3 such products, the elements of the Klein 4 group.

You can use the above counting technique to find the cardinality of the class of any product of cycles in S_n. The argument just gets a little more involved.