the given problem is simply a complement of MDFA having 3 consecutive 1’s over the input $(0,1).$ We can take complement by simply changing the non-final state to the final state and vice-versa.
The MDFA not containing 3 or more consecutive 1’s will require $4$ state out of which $3$ are final and $1$ is dead states.
it will accept strings like $(\epsilon,0,1,00,01,10,11,000,001,010,011,100,101,110...\infty)$
it will reject strings like $(111,1110,11101,11111,111000,1110101,111001...\infty)$
the correct MDFA is as follows:
The correct answer is $4$