As far as for above problem say regular expression for $(0+1)(0+1)\ldots 3$ times $=(0+1)(0+1)(0+1)$
Having DFA:
So, for regular expression $(0+1)(0+1)...N$ times we have $\mathbf{N+2}$ states in DFA
$N+1$ states in NFA ( we can remove dead state)
When question is about minimum state finite automata (and nothing is mentioned NFA/DFA) then which ever having minimum number must be taken which here is $N+1$ states.
Correct Answer: $B$