This becomes true at $a = \frac{1}{3}$.
Claim. If $G$ is an $n$-vertex bipartite graph such that $\delta(G) \geq \frac{1}{3}n$, then $\delta(G)=\kappa(G)$.
Proof. Let $(A,B)$ be the bipartition of $G$. Suppose $\kappa(G)<\delta(G)$ and let $X \subseteq V(G)$ be such that $G-X$ is disconnected and $|X|=\kappa(G)$. Let $A=A_1 \sqcup A_2 \sqcup A_3$ and $B=B_1 \sqcup B_2 \sqcup B_3$ be such that $A_1=X \cap A$, $B_1=X \cap B$, and $G[A_2 \cup B_2]$ and $G[A_3 \cup B_3]$ are both a union of connected components of $G-X$. Since $\kappa(G) < \delta(G)$, none of $A_2, A_3, B_2, B_3$ are empty. Observe that the degree in $G$ of each vertex in $A_2, A_3, B_2, B_3$ are at most $|B_2|+|B_1|, |B_3|+|B_1|, |A_2|+|A_1|, |A_3|+|A_1|$, respectively. By averaging, some vertex in $A_2 \cup A_3 \cup B_2 \cup B_3$ has degree at most $(n+\kappa(G))/4$. Thus $\delta(G) \leq (n+\kappa(G))/4 < (n+\delta(G))/4$, which implies $\delta(G) < \frac{n}{3}$.
On the other hand, the following claim shows that $\frac{1}{3}$ is best possible.
Claim. For all $\ell \in \mathbb{N}$ exists a bipartite graph $G$ on $3\ell+1$ vertices such that $\delta(G)=\ell$, but $\kappa(G) < \delta(G)$.
Proof. Let $G_1$ and $G_2$ be copies of $K_{\ell, \ell}$ with bipartitions $(A_1,B_1)$ and $(A_2,B_2)$, respectively. Let $X_1 \subseteq A_1$ and $X_2 \subseteq A_2$ both be of size $k$, and let $G$ be the graph obtained from $G_1$ and $G_2$ by identifying $X_1$ and $X_2$. Note that $G$ is a bipartite graph with $4\ell-k$ vertices, minimum degree $\ell$, and vertex-connectivity $k$. Setting $k=\ell-1$ proves the claim.