Variance of an observation of an arbitrary state measured consecutively by two operators

Since $A$ and $B$ are commuting observables, we can express a general state $|Psi rangle$ as a superposition of simultaneous eigenstates of $A$ and $B$. In other words, we can say that begin{equation} | Psi rangle = sum_{a,b} c_{a,b} |arangle | b rangle end{equation} where $c_{a,b}$ are complex coefficients, $|arangle$ is an eigenstate of $A$ with eigenvalue $a$ ($A | a rangle = a | a rangle$), and $|brangle$ is an eigenstate of $B$ with eigenvalue $b$ ($B | b rangle = b | b rangle$). I'll take the states to be normalized so that $langle a | a' rangle =delta_{a,a'}$ and $langle b | b' rangle = delta_{b,b'}$.

The variance of $B$ is

begin{eqnarray} sigma_B^2 &=& langle Psi | B^2 | Psi rangle - |langle Psi | B | Psi rangle |^2 nonumber \ &=& sum_{a,b} b^2 |c_{a,b}|^2 - left|sum_{a,b} b |c_{a,b}|^2 right|^2 end{eqnarray}

If you measured $A$ to have a definite value $a_star$, then we can define $c_b equiv c_{a_star,b}$, and the above expression becomes

begin{equation} sigma_B^2 = sum_{b} b^2 |c_{b}|^2 - left|sum_{b} b |c_{b}|^2 right|^2 end{equation}

That's the best you can do without knowing the state exactly. One important special case is if the system is in an eigenstate of $B$. Then $|c_b|=1$ and the variance is zero.