$\text{1. Every relation in 3NF is also in BCNF}$

For a $3NF$ relation to be in $BCNF$, that has to has satisfied this condition-

$\text{for each dependency }$ $X \rightarrow Y$, $\text{X should be the super key.}$

So, this statement is wrong

$\text{2. A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent}$ $\text{on every key of R}$

This is a straightforward definition of $2NF$. If a relation is in $1NF$ and in that relation each & every non-prime attributes has to be fully functionally dependent on each key of the relation.

& for a $2NF$ relation to be in $3NF$, that has to satisfy the condition which states, no non-prime attributes should be transitively dependent on candidate key or we can say that, no non-prime attribute should be determined by anything but a super key

Let's take an example:

$R(ABCDE)$ and $F = \{BC \rightarrow ADE, A \rightarrow BCDE, D \rightarrow E\}$

The candidate keys will be = $\{ A, BC\}$

Prime Attribute = $A,B,C$

Non-Prime Attribute = $D,E$

Here, all the non-prime attributes$(D,E)$ are fully functionally dependent on each key of $R$

But here, $E$(non-prime attribute) can be determined by $D$ and $D$ is not the candidate key. So, $E$ is transitively dependent on the candidate keys of $R$ and moreover, $D$ is not the super key also.

$\therefore$ This relation is not in $3NF$. For making it to $3NF$ we have to break the relation into $2$ parts: $R_1(ABCD), F=\{ \{A \rightarrow BCD\}, BC \rightarrow AD\}$ & $R_2(DE), F = \{D \rightarrow E\}$

This option is also wrong.

$\text{3. Every relation in BCNF is also in 3NF}$

Obviously, For a relation to be in $BCNF$, it has to be in $3NF$ first.

$\color{Green}{\therefore \text{This statement is Correct}}$

$\text{4. No relation can be in both BCNF and 3NF}$

Yes, there can be some relation which is in $3NF$ and as well as $BCNF$

and if a relation is in $BCNF$ then it is also in $3NF$

So, this statement is also wrong.

Correct option is $C)$