In DFA, for every state we have transition for each symbol who's belongs to given alphabet.
Alphabet Set(Σ)= {0,1}
So, in given NFA, for q0 state, 2 transitions are there for 0 and 1. // (q0, 0)=q0 and (q0,1)=q1
for q1 state, 1 transition is there for symbol 1. // (q1, 1)=q2
for q2 state, 1 transition is there for symbol 0. // (q2, 0)=q1
Now, left transitions are like (q1, 0), (q2, 1) belongs to dead state lets say q3