Since it is a DFA, exactly 2 transitions should be there for each state.
From state q0, possible transitions on a are $\delta (q0,a) \rightarrow q0$ and $\delta (q0,a) \rightarrow q1$
From state q0, possible transitions on b are $\delta (q0,b) \rightarrow q0$ and $\delta (q0,b) \rightarrow q1$
From state q1, possible transitions on a are $\delta (q1,a) \rightarrow q0$ and $\delta (q1,a) \rightarrow q1$
From state q1, possible transitions on b are $\delta (q1,b) \rightarrow q0$ and $\delta (q1,b) \rightarrow q1$
So, 2*2*2*2 = 16 DFAs possible
