If $(B,\cup,\cap,\lnot,0,1)$ is a Boolean algebra, the Boolean ring is just the same algebra with $\cap,0,$ and $1$ corresponding to $\times,0,$ and $1$, respectively, in the ring, and $x+y = (x\cap \lnot y)\cup(\lnot x\cap y)$. From this view, the more useful thing to look at is the "co-atoms" - the elements of the Boolean algebra whose complements are atoms.
That's because, when we consider a Boolean algebra as a ring, the ideals of the ring turn out to be sets $I$ such that, if $x,y\in I$ then $x\cup y\in I$, and, if $x\in I$, $y$ any element, $x\cap y\in I$. That such sets are ideals is easy to show. To prove that all ideals of the Boolean ring satisfy this property, you need only prove the following formula:
$$x + y + (x\cap y) = x\cup y$$
If $x$ is any element, the ring ideal, $\left<x\right>$, generated by $x$, is the set of $y$ such that $x\cap y = y$.
Now, it turns out that $x$ is a co-atom if and only if $\left<x\right>$ is a maximal ideal of the ring.
So $a$ is an atom of your Boolean algebra if and only if $\left<\lnot a\right>$ is a maximal ideal of the Boolean ring. Then $B/\left<\lnot a\right>\cong \mathbb Z/\left<2\right>$.
Not all maximal ideals in a boolean algebra necessarily correspond to atoms, however.