Let $R$ be a ring, $C, A$ two $R$-modules. For all $x\in\operatorname{Ext}_R^1(C,A)$ I have to construct a short exact sequence $$0\to A\to B\to C\to 0$$ of $R$-modules such that $\partial(id_A)=x$, where $\partial:\operatorname{Hom}(A,A) \to \operatorname{Ext}_R^1(C,A)$ comes from the induced long exact sequence $$\dots\leftarrow\operatorname{Ext}_R^1(C,A)\leftarrow \operatorname{Hom}(A,A) \leftarrow \operatorname{Hom}(B,A)\leftarrow \operatorname{Hom}(C,A) \leftarrow 0.$$
I have no idea how to prove this, sorry. Could you give me a hint? I appreciate your help.
The standard construction is via a push-out.
Consider an exact sequence $0\to K\to P\to C\to 0$, with $P$ projective. Applying $\operatorname{Hom}_R(-,A)$, we get the exact sequence $$\def\H{\operatorname{Hom}_R}\def\E{\operatorname{Ext}_R^1} 0\to\H(C,A)\to\H(P,A)\to\H(K,A)\xrightarrow{\delta}\E(C,A)\to0 $$ Take $f\colon K\to A$ such that $\delta(f)=x$, so you can build the push-out diagram $$\require{AMScd} \begin{CD} 0 @>>> K @>>> P @>>> C @>>> 0 \\ @. @VfVV @VVV @| \\ 0 @>>> A @>>> B @>>> C @>>> 0 \end{CD} $$
Your task is now to prove that the exact sequence in the bottom row doesn't depend on $f$, but only on $x$ and that this is indeed the exact sequence you were looking for.
Note that $\E(C,A)$ is exactly the cokernel of $\H(P,A)\to\H(K,A)$, essentially by definition, so $\delta$ is nothing else than the canonical projection.