Total numbers of 3 -digit numbers that are divisible by 6 and can...

A number is divisible by $6$ when it is divisible by $2$ and $3$ both.

The three digit number can be $\mathop{a}_{¯}\mathop{b}_{¯}\mathop{2}_{¯}$.

$a + b$ can be $4, 7, 10$

If $a + b = 4$ then $( a, b )$ can be $( 1, 3 )$, $( 3, 1 )$, $( 2, 2 )$

If $a + b = 7$ then $( a, b )$ can be$( 2, 5 ) , ( 5, 2 ) , ( 3, 4 ) , ( 4, 3 )$

If $a + b = 10$ then $( a, b )$ can be $( 5, 5 )$

So, $8$ such cases when $2$ is at unit's place

Similarly, there exist $8$ such cases when $4$ is at unit's place.

Total $= 16$ cases.

Hence this is the required answer.