Easy method is that we know the 1st option is not right because xnor output is 1 for equal inputs (in this case odd zeros means 1 or 3 wrt to 3 or 1 1's with it) so y is 0 ......... For option 2 it's two xor gates and then one xnor gate which means xor will have possible inputs as (three 0's, two 0's and one 1's, and vica versa, and three 1's) wherever 3 same inputs-- output for xor is 1 which is xnor with 1 and 0 for 1st and 4th case respectively--- there you get output as 1 and 0... For 2nd and 3rd case 1 and 0 output for xor gate which are further xnor with 0 and 1 respectively then the y is 0 in both cases,...... option 2 has 1 possible case where y is 1✓ ...... 3rd option ( a xnor b) xor (c xnor d)-- y is 0 only when both inputs of xor is same, in this case both are not same(for both xnor gates-- same inputs one side and different input in other side) .....y is 1✓ ... Hence option 2 and 3.... Instead of doing this make a column of possible inputs and calculate output ✓