We know that any permutation in $S_n$ is a product of disjoint cycles. And order of an $r$ cycle is $r$. The order of any permutation is lcm of the order of all $r$-cycles in it. We shall require these things.
First of all, if we need to search element of order 6 in $S_4$ then it is (by the above logics) either a 6 cycle or disjoint product of 2-cylce and 3-cycle. Why? because in the first case then order is directly becoming 6 and for the second case, the order is lcm$(2, 3)=6$.
But since $S_4$ is nothing but group of all permutations on 4 distinct symbols, so using 4 symbols we can't create 6-cycle. First case is thus cancel out. For the second case, if we choose any 3-cycle then we are left with 1 symbols using which we can't create any 2-cycle.
Hence no element of order 6 is there in $S_4$.