Using the closure set of attributes we get the primary key as $F_{1}.$
Since, there is only one candidate key and it has only one attribute we cannot have a partial functional dependency. So, the table must be in $2\;NF.$
We have $F_2 \to F_3,$ where $F_2$ is not a prime attribute and hence this violates $3\;NF$ requirement. So, correct answer is B.