Let $B$ be a Boolean algebra and $A$ be its subalgebra. The claim is the variety generated by $A$ is the class of all Boolean algebras. I can give the following proof for it but I think this claim is not true. Please let me know if this proof is true. If that is false please show me the mistake.
Let $B_2$ be the Boolean algebra with two elements. Then $V(B_2)$ is the set of all Boolean algebras, as it is discussed here by the following argument:
Every variety is generated by is subdirectly irreducible members.
$B_2$ is the only subdirectly irreducible member of the variety axiomatized by the identities of Boolean algebra.
Now we know that $B_2$ is a subalgebra of $A$ (my doubt is here but I can not refuse that) and so $B_2\in V(A)$. Therefore $V(B_2)\subseteq V(A)$. But, since $V(B_2)$ is the set of all Boolean algebras, we conclude that $V(A)$ is also the set of all Boolean algebras.
To make it more clarify why I think that this can't be true is that in the page 110 of the paper "Pretabular varieties of modal algebras" of Block, $A$ is a subalgebra of a Boolean algebra $\mathcal{F}^+$ but he uses the variety $V(A)$ without mentioning that $V(A)$ is the class of all Boolean algebras.