The last digit of $7^k$ repeats as $7$, $9$, $3$, $1$, $7$, $\cdots$. Therefore $7^{23}$ ends with $3$.

The last digit of $8^k$ repeats as $8$, $4$, $2$, $6$, $8$, $\cdots$. Therefore $8^{105}$ ends with $8$.

The last digit of $3^k$ repeats as $3$, $9$, $7$, $1$, $3$, $\cdots$. Therefore $3^{18}$ ends with $9$.

Hence, the units digit of $7^{23} \times 8^{105} \times 3^{18}$ is the same as the last digit of $3\times 8\times 9$ which is $6$.