Short method:
Johnson’s counter is Mod 2N counter and ring is Mod N counter.
Detailed Explanation:
If you look closely at the question, you will notice that the input of J and K are complements of each other and if you recall, if the input values are the complement of each other, JK FF behaves as D FF.
Here in the diagram we can see how D FF is formed from JK FF.
So this means the JK in our question will behave as D FF. You can also replace the JK FF with D FF if you want for easier understanding.
So now we can equate the values of J with Q,
Hence we can get the state diagram. And we will have 6 states since it was mentioned that it will start from 000 thereby eliminating the possibility of MOD 2 counter.
Also the shift register option is correct.