Let us work over $mathbb{C}$.
The inclusion $u colon B to A$ induces a surjection $hat{u} colon A^{vee} to B^{vee}$.
By general facts on Abelian varieties, the kernels of $u$ and $hat{u}$ have the same number of connected components. Since $u$ is injective, its kernel is trivial, so it follows $ker hat{u}=(ker hat{u})_0$; in other words $ker hat{u}$ is an Abelian subvariety of $A^{vee}$.
Therefore we have an exact sequence of Abelian varieties
$$0 to ker hat{u} to A^{vee} to B^{vee} to 0.$$ By dualizing it, we obtain $$0 to B to A to (ker hat{u})^{vee} to 0,$$
that is $C = (ker hat{u})^{vee}$.
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