Here my solution: Suppose there exists and $H \leq A_4 \oplus Z_3$ such that order of H is 18. Now, notice index of H in $A_4 \oplus Z_3$ is 2. therefore, H is normal, and therefore, the $A_4 \oplus Z_3 / H$ exists. Now, for every $\alpha$ in $A_4 \oplus Z_3$, $\alpha^2H = (\alpha H)^2 = H$ since the quotient group has order 2. But notice, $A_4$ has only 3 elements of order 2, therefore, $A_4 \oplus Z_3$ have only 3 elements of order 2, which implies H has 3 elements of order 2 and the rest must have order 1, and this is an absurd.
Is this a correct solution? Do you guys have any other solution?