If a number is divisible by both 2 and 3, then it will be divisible by 6 also.
Divisibility by 2
Here all the numbers, 9638, 9640, 9642, 9648 are divisible by 2 as they end with one of these (0, 2, 4, 6, 8) numbers or simply as they are even numbers.
Divisibility by 3
$\Rightarrow$9 + 6 + 3 + 8 = 26
$\Rightarrow$9 + 6 + 4 + 0 = 19
$\Rightarrow$9 + 6 + 4 + 2 = 21
$\Rightarrow$9 + 6 + 4 + 8 = 27
Here only 21 and 27 are divisible by 3 hence only 9642 and 9648 are divisible by 3.
So, these two numbers are divisible by both 2 and 3 and hence divisible by 6.
Divisibility by 8
A number will be divisible by 8 only if the last three digits are divisible by 8.
The last 3 digits of only 9648, i.e. 648 is divisible by 8.
But last three digits of 9642, i.e. 642 is not divisible by 8.
 9648 is divisible by both 6 and 8.