For $c_1$ the problem is solved. $\newcommand{\bZ}{\mathbb{Z}}$ For any smooth manifold and any $c\in H^2(M,\bZ)$ there exists a smooth complex line bundle $L\to M$ such that $c_1(L)=c$.

By results of Thom, for any oriented manifold $M$, any $\alpha\in H_{n-4}(M,\bZ)$ is represented by an oriented submanifold.

On the other hand, for any $n\geq 7$, there exists an $n$-dimensional oriented manifold $M$ and a homology class $\alpha\in H_{n-4}(M,\bZ)$ such that the normal bundle of any submanifold representing $\alpha$ does not admit a $spin^c$-structure; see Theorem 3, page 9 of this paper.

If $\alpha^\dagger\in H^4(M,\bZ)$ denotes the Poincare dual of such an $\alpha$, then there exist no rank 2 complex vector bundle $E\to M$ such that $c_2(E)=\alpha^\dagger$.

If such a bundle existed, then the zero set of a generic section of $E$ will be an oriented submanifold $S$ of $M$ representing $\alpha$. The normal bundle of $S$ in $M$ is isomorphic to $E|_S$. In particular it admits $spin^c$ structures because it admits an almost complex structure.

Edit 1. A rather deep divisibility theorem shows that if $n\geq 2$ and $E\to S^{2n}$ is a rank $n$ complex vector bundle, then $c_n(E)\in H^{2n}(S^{2n},\bZ)$ is divisible by $n!$.

For $c_1$ the problem is solved. $\newcommand{\bZ}{\mathbb{Z}}$ For any smooth manifold and any $c\in H^2(M,\bZ)$ there exists a smooth complex line bundle $L\to M$ such that $c_1(L)=c$.

By results of Thom, for any oriented manifold $M$, any $\alpha\in H_{n-4}(M,\bZ)$ is represented by an oriented submanifold.

On the other hand, for any $n\geq 7$, there exists an $n$-dimensional oriented manifold $M$ and a homology class $\alpha\in H_{n-4}(M,\bZ)$ such that the normal bundle of any submanifold representing $\alpha$ does not admit a $spin^c$-structure; see Theorem 3, page 9 of this paper.

If $\alpha^\dagger\in H^4(M,\bZ)$ denotes the Poincare dual of such an $\alpha$, then there exist no rank 2 complex vector bundle $E\to M$ such that $c_2(E)=\alpha^\dagger$.

If such a bundle existed, then the zero set of a generic section of $E$ will be an oriented submanifold $S$ of $M$ representing $\alpha$. The normal bundle of $S$ in $M$ is isomorphic to $E|_S$. In particular it admits $spin^c$ structures because it admits an almost complex structure.

Edit 1. A rather deep divisibility theorem shows that if $n\geq 3$ and $E\to S^{2n}$ is a complex vector bundle, then $c_n(E)\in H^{2n}(S^{2n},\bZ)$ is divisible by $(n-1)!$.

For $c_1$ the problem is solved. $\newcommand{\bZ}{\mathbb{Z}}$ For any smooth manifold and any $c\in H^2(M,\bZ)$ there exists a smooth complex line bundle $L\to M$ such that $c_1(L)=c$.

By results of Thom, for any oriented manifold $M$, any $\alpha\in H_{n-4}(M,\bZ)$ is represented by an oriented submanifold.

On the other hand, for any $n\geq 7$, there exists an $n$-dimensional oriented manifold $M$ and a homology class $\alpha\in H_{n-4}(M,\bZ)$ such that the normal bundle of any submanifold representing $\alpha$ does not admit a $spin^c$-structure; see Theorem 3, page 9 of this paper.

If $\alpha^\dagger\in H^4(M,\bZ)$ denotes the Poincare dual of such an $\alpha$, then there exist no rank 2 complex vector bundle $E\to M$ such that $c_2(E)=\alpha^\dagger$.

If such a bundle existed, then the zero set of a generic section of $E$ will be an oriented submanifold $S$ of $M$ representing $\alpha$. The normal bundle of $S$ in $M$ is isomorphic to $E|_S$. In particular it admits $spin^c$ structures because it admits an almost complex structure.

For $c_1$ the problem is solved. $\newcommand{\bZ}{\mathbb{Z}}$ For any smooth manifold and any $c\in H^2(M,\bZ)$ there exists a smooth complex line bundle $L\to M$ such that $c_1(L)=c$.

By results of Thom, for any oriented manifold $M$, any $\alpha\in H_{n-4}(M,\bZ)$ is represented by an oriented submanifold.

On the other hand, for any $n\geq 7$, there exists an $n$-dimensional oriented manifold $M$ and a homology class $\alpha\in H_{n-4}(M,\bZ)$ such that the normal bundle of any submanifold representing $\alpha$ does not admit a $spin^c$-structure; see Theorem 3, page 9 of this paper.

If $\alpha^\dagger\in H^4(M,\bZ)$ denotes the Poincare dual of such an $\alpha$, then there exist no rank 2 complex vector bundle $E\to M$ such that $c_2(E)=\alpha^\dagger$.

If such a bundle existed, then the zero set of a generic section of $E$ will be an oriented submanifold $S$ of $M$ representing $\alpha$. The normal bundle of $S$ in $M$ is isomorphic to $E|_S$. In particular it admits $spin^c$ structures because it admits an almost complex structure.

Edit 1. A rather deep divisibility theorem shows that if $n\geq 2$ and $E\to S^{2n}$ is a rank $n$ complex vector bundle, then $c_n(E)\in H^{2n}(S^{2n},\bZ)$ is divisible by $n!$.