There is no canonical way to distinguish a particular theta divisor within its algebraic equivalence class. So, in order to get a meaningful answer, one must specify which theta divisor is actually choosing.

For instance, the following result can be found in [Birkhenake-Lange, Complex Abelian Varieties, Exercise 10 p. 361].

Proposition. Let $C$ be a smooth curve of genus $g$ and let $\alpha_c \colon C \to J$ be the embedding with respect to the point $c \in C$.

Let $\Theta \subset J$ be the theta divisor defined by $$\alpha_L^* \Theta = W_{g-1},$$
where $L=\omega_g \otimes \mathscr{O}((1-g) \cdot c)$. Then one has $$\alpha_c^* \mathscr{O}_J(\Theta) = \mathscr{O}_C (g \cdot c).$$

There is no canonical way to distinguish a particular theta divisor within its algebraic equivalence class. So, in order to get a meaningful answer to any question of this kind, one must specify which theta divisor is actually considering.

For some particular choice of $\Theta$, the answer can be found in [Birkhenake-Lange, Complex Abelian Varieties, Exercise 10 p. 361]. For the reader's convenience I will restate it here.

Proposition. Let $C$ be a smooth curve of genus $g$ and let $\alpha_c \colon C \to J$ be the embedding with respect to the point $c \in C$.

Let $\Theta \subset J$ be the theta divisor defined by $$\alpha_L^* \Theta = W_{g-1},$$
where $L=\omega_g \otimes \mathscr{O}((1-g) \cdot c)$. Then one has $$\alpha_c^* \mathscr{O}_J(\Theta) = \mathscr{O}_C (g \cdot c).$$