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The answer is no if one makes the additional assumption that $S \in |-K_V|$ is ample, i.e. that $V$ is a smooth Fano threefold, as shown by the following example.

If $V$ is a Fano 3-fold containing $S$ as an ample divisor, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective (with torsion free cokernel).

Now, it is well-known that for all $g \geq 2$ there exists a polarized $K3$ surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $L$, see for instance

Picard group of a K3 surface generated by a curve

Therefore the injectivity of $r$ implies $\rho(V)=1$, or equivalently $b_2(V)=1$.

On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).

Therefore, if $g$ is large enough (it sufficies to take $g \geq 13$) we necessarily have $b_2(V) \geq 2$, a contradiction.

Remark 1. A. Beauville proved in this paper that a general $K3$ surface with given Picard lattice $R$ and polarization class $L \in R$ is an anticanonical divisor in a smooth Fano threefold if and only if there exists an isomorphism of polarized lattices $(R, \, L) = (\textrm{Pic}(V), \, K^{−1}_V)$ for some smooth Fano threefold $V$.

Remark 2. As observed in the comments below by D. Litt, J. C. Ottem and Mark, it is actually possible that a $K3$ surface $S$ appears as a non-ample anticanonical divisor in a smooth 3-fold, so the question in its general form is still unanswered.

The answer is no if one makes the additional assumption that $S \in |-K_V|$ is ample, i.e. that $V$ is a smooth Fano threefold, as shown by the following example.

If $V$ is a Fano 3-fold containing $S$ as an ample divisor, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective (with torsion free cokernel).

Now, it is well-known that for all $g \geq 2$ there exists a polarized $K3$ surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $L$, see for instance

Picard group of a K3 surface generated by a curve

Therefore the injectivity of $r$ implies $\rho(V)=1$, or equivalently $b_2(V)=1$.

On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).

Therefore, if $g$ is large enough (it sufficies to take $g \geq 13$) we necessarily have $b_2(V) \geq 2$, a contradiction.

Remark 1. A. Beauville proved in this paper that a general $K3$ surface with given Picard lattice $R$ and polarization class $L \in R$ is an anticanonical divisor in a smooth Fano threefold if and only if there exists an isomorphism of polarized lattices $(R, \, L) = (\textrm{Pic}(V), \, K^{−1}_V)$ for some smooth Fano threefold $V$.

Remark 2. As observed in the comments below by D. Litt, J. C. Ottem and Mark, it is actually possible that a $K3$ surface $S$ appears as a non-ample anticanonical divisor in a smooth 3-fold, so the question in its general form is still unanswered.

The answer is no if one makes the additional assumption that $S \in |-K_V|$ is ample, i.e. that $V$ is a smooth Fano threefold, as shown by the following example.

If $V$ is a Fano 3-fold containing $S$ as an ample divisor, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective with torsion free cokernel.

Now, it is well-known that for all $g \geq 2$ there exists a polarized $K3$ surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $L$, see for instance

Picard group of a K3 surface generated by a curve

Therefore the injectivity of $r$ implies $\rho(V)=1$, or equivalently $b_2(V)=1$.

On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).

Therefore, if $g$ is large enough (it sufficies to take $g \geq 13$) we necessarily have $b_2(V) \geq 2$, a contradiction.

Remark 1. A. Beauville proved in this paper that a general $K3$ surface with given Picard lattice $R$ and polarization class $L \in R$ is an anticanonical divisor in a smooth Fano threefold if and only if there exists an isomorphism of polarized lattices $(R, \, L) = (\textrm{Pic}(V), \, K^{−1}_V)$ for some smooth Fano threefold $V$.

Remark 2. As remarked in the comments below by D. Litt, J. C. Ottem and Mark, it is actually possible that a $K3$ surface $S$ appears as a non-ample anticanonical divisor in a smooth 3-fold, so the question in its general form is still unanswered.

The answer is no if one makes the additional assumption that $S \in |-K_V|$ is ample, i.e. that $V$ is a smooth Fano threefold, as shown by the following example.

If $V$ is a Fano 3-fold containing $S$ as an ample divisor, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective (with torsion free cokernel).

Now, it is well-known that for all $g \geq 2$ there exists a polarized $K3$ surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $L$, see for instance

Picard group of a K3 surface generated by a curve

Therefore the injectivity of $r$ implies $\rho(V)=1$, or equivalently $b_2(V)=1$.

On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).

Therefore, if $g$ is large enough (it sufficies to take $g \geq 13$) we necessarily have $b_2(V) \geq 2$, a contradiction.

Remark 1. A. Beauville proved in this paper that a general $K3$ surface with given Picard lattice $R$ and polarization class $L \in R$ is an anticanonical divisor in a smooth Fano threefold if and only if there exists an isomorphism of polarized lattices $(R, \, L) = (\textrm{Pic}(V), \, K^{−1}_V)$ for some smooth Fano threefold $V$.

Remark 2. As observed in the comments below by D. Litt, J. C. Ottem and Mark, it is actually possible that a $K3$ surface $S$ appears as a non-ample anticanonical divisor in a smooth 3-fold, so the question in its general form is still unanswered.

The answer is no under the additional assumption that the anticanonical system of the $3$-fold $V$ is ample, i. e. that $V$ is Fano (in a precedent version of the answer I wrongly assumed that this was implicit in the question, see the comments below).

By the sake of simplicity let us assume that $-K_V$ is very ample, for the general case you can look at this paper by A. Beauville.

If $V$ is a Fano 3-fold containing $S$ as a hyperplane section, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective, and this is a constraint on the $K3$ surface $S$.

For instance, it is well-known that for all $g \geq 2$ there exists a polarized K3 surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $[L]$, see for instance

Picard group of a K3 surface generated by a curve

On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).

Therefore, if $g$ is large enough, every smooth Fano $3$-fold $V$ of genus $g$ satisfies $b_2(V) \geq 2$, i.e. $\textrm{Pic}\, V $ has rank $\geq 2$. Since $\textrm{Pic} \, S$ has rank $1$, by the injectivity of $r$ it follows that $S$ cannot be obtained as a hyperplane section of a smooth Fano $3$-fold.

Let me finish by asking the following

Question. Is there any example of a smooth $K3$ surface $S$ appearing as a non-ample anticanonical divisor in a smooth 3-fold $V$?

The answer is no if one makes the additional assumption that $S \in |-K_V|$ is ample, i.e. that $V$ is a smooth Fano threefold, as shown by the following example.

If $V$ is a Fano 3-fold containing $S$ as an ample divisor, by the Lefschetz Hyperplane Theorem the restriction map $$r \colon \textrm{Pic}\,V \longrightarrow \textrm{Pic} \, S$$ is injective with torsion free cokernel.

Now, it is well-known that for all $g \geq 2$ there exists a polarized $K3$ surface $(S, \, L)$ of genus $g$ such that $\textrm{Pic} \, S$ is generated by $L$, see for instance

Picard group of a K3 surface generated by a curve

Therefore the injectivity of $r$ implies $\rho(V)=1$, or equivalently $b_2(V)=1$.

On the other hand, we know by the work of Iskovskih that smooth Fano threefolds with $b_2=1$ form a bounded family (they belong to exactly $17$ isomorphism classes).

Therefore, if $g$ is large enough (it sufficies to take $g \geq 13$) we necessarily have $b_2(V) \geq 2$, a contradiction.

Remark 1. A. Beauville proved in this paper that a general $K3$ surface with given Picard lattice $R$ and polarization class $L \in R$ is an anticanonical divisor in a smooth Fano threefold if and only if there exists an isomorphism of polarized lattices $(R, \, L) = (\textrm{Pic}(V), \, K^{−1}_V)$ for some smooth Fano threefold $V$.

Remark 2. As remarked in the comments below by D. Litt, J. C. Ottem and Mark, it is actually possible that a $K3$ surface $S$ appears as a non-ample anticanonical divisor in a smooth 3-fold, so the question in its general form is still unanswered.