for a state, on an input symbol we have two choices, which are
- end up in the same state
- end up in the other state
There are two input symbols available.
So, total number of ways transition is possible from a state = $\text{Choices for an input symbol} \times \text{Number of Input Symbols} = 2\times 2 = 4$
Total number of ways in which transition can happen in a DFA with two states = $\text{Choices at state }q_0 \times \text{Choices at state }q_1 = 4 \times 4 =16$
In a DFA any state can be a final state or an initial state. Here, Input state is fixed, so we need not worry about it.
total number of possibilities for the set of final states = $4$
Hence, total states in this DFA = $4 \times 16 = 64$