I think it should be option B.
Just a clear intuition, whenever we see a 1 we are changing the state.
For 1st ‘1’ we are writing ‘O’, for 2nd ‘1’ we are writing ‘E’, for 3rd ‘1’ again ‘O’ and so on.
So basically we would be able to know the number of 1’s encountered even or odd at any point.
Answer must be option B i guess as it is a mealey machine it will always give a output to every single input symbol.Testing strings of different length and different count of both 1’s and 0’s each time it can be concluded that the last output given by the machine is whether the the count of 1’s in the string is even or odd.