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 $C = (\ker \hat{u})^{\vee}$.

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 $C = (\ker \hat{u})^{\vee}$.