Let $\mathcal{C}$ be an abelian category. In $\mathcal{C}$ we consider the diagram \begin{array}{ccc} A&&\\\ \downarrow&&\\\ C&\rightarrow&D \end{array} with arrows being monomorphism.
Is it possible to say when there is a $B$ such that we have a push-out square \begin{array}{ccc} A&\rightarrow&B\\\ \downarrow&&\downarrow\\\ C&\rightarrow&D \end{array} and if possible how to get $B$ (up to isomorphism)?
If there is such a pushout, then $B\to D$ is also a monomorphism, i.e. $B$ is a subobject of $D$.
Phrased more concretely, you're asking when there is a subobject $B$ of $D$ such that $B+C = D$ and $B\cap C = A$.
Let us mod out $D$ by $A$: we get a monomorphism $C/A\to D/A$, and a pushout square
$\require{AMScd}\begin{CD} C @>>> D \\ @VVV @VVV \\ C/A @>>> D/A\end{CD}$
which we can glue to the previous one to get a pushout square
$\require{AMScd}\begin{CD} A @>>> B \\ @VVV @VVV \\ C/A @>>> D/A\end{CD}$
This means that $D/A\cong C/A\oplus B/A$
Conversely, assume $C/A \to D/A$ has a summand $X$, and consider its pullback $B$ to $D$. Then clearly $B\cap C = A$ and $B+C = D$.
So there is such a $B$ if and only if $C/A$ has a summand in $D/A$; and in fact if you're more careful with the above constructions, it seems like
$\{$ subobjects $B\to D$ realizing such a pushout $\}\to \{$ summands of $C/A$ in $D/A \}$
is a bijection, with inverse given by the pullback; this can be seen as a relativization of the claim that subjects of $D$ containing $A$ are the same thing as subobjects of $D/A$.
