jmc

There are $10$ numbers with a $6$ in the hundreds place and a $6$ in the units place.

There are $10$ numbers with a $6$ in the hundreds place and a $6$ in the tens place.

There are $8$ numbers with a $6$ in the tens place and a $6$ in the units place. (Remember, we only have the numbers from $1$ to $800.$)

Adding them up, we have $10+10+8=28.$ However, we counted $666$ three times. Therefore subtract $2$ from $28$ to get $\boxed{26}$ numbers containing the digit $6$ at least twice.