Number of a is divisible by 2 but not by 3.

So from all the multiples of 2 we have to exclude the multiples of 3.

Note that common multiplier of 2 & 3 can be found in the table of 6 bcz LCM(2,3)=6.

SO THE NUMBER DIVIDED BY 6 WE HAVE TO EXCLUDE LIKE 6,12,18,24,... FROM TABLE OF 2.

now the problem reduce to

Complement of ( min. DFA which accepts number of "a" which is divisible by 6)

Now to design this DFA we need 6 states minimum.

After designing the DFA inside the bracket have to complement it by changing the final states.