Initial Case Applying a mole balance to the single CSTR in steady state yields
\begin{align}
v_0C_{A0} - v_0C_{A} - kC_{A}V &= 0 \\
C_{A} &= \frac{v_0C_{A0}}{v_0 + kV} \tag{1}
\end{align}
The conversion, combining with Eq. (1), is
\begin{align}
X_1 &= \frac{C_{A0} - C_A}{C_{A0}} \\
X_1 &= \frac{C_{A0} - \dfrac{v_0C_{A0}}{v_0 + kV}}{C_{A0}} \\
X_1 &= 1 - \frac{v_0}{v_0 + kV} \\
X_1 &= \frac{kV}{v_0 + kV} \rightarrow kV = \frac{X_1v_0}{1 - X_1} \tag{2}
\end{align}
We will use Eq. (2) later.
Second Case Now the volumetric flow rate splits in two. Let me call that value in the upper branch $ \alpha v_0 $, and the one for the lower branch $(1 - \alpha)v_0$. Since we have a bifurcation, the concentration of reactant $A$ doesn't change, so the value is still $C_{A0}$ when entering both reactors.
We denote the exit concentration in the reactor at the top $C_{A1}$, and at the bottom $C_{A2}$. Applying a mole balance to both CSTR's in steady state, with equal volumes but of value $V/2$, yields
\begin{align}
\alpha v_0C_{A0} - \alpha v_0 C_{A1} &- (k C_{A1})\frac{V}{2} = 0 \\
(1 - \alpha)v_0C_{A0} - (1 - \alpha)v_0C_{A2} &- (kC_{A2})\frac{V}{2} = 0 \\
C_{A1} = \frac{\alpha v_0C_{A0}}{\alpha v_0 + kV/2} &\hspace{0.25 cm}
C_{A2} = \frac{(1 - \alpha)v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{3,4} \\
\end{align}
Now, upon mixing the streams, the volumetric flow rate recovers its value $v_0$. However, the concentration changes, and we denote its value by $C_{A3}$. This one is obtained by a mole balance at steady state, combining with Eqs. (3) and (4)
\begin{align}
\alpha v_0C_{A1} + (1 - \alpha)v_0C_{A2} &= v_0 C_{A3} \\
C_{A3} &= \alpha C_{A1} + (1 - \alpha)C_{A2} \\
C_{A3} &= \frac{\alpha^2 v_0C_{A0}}{\alpha v_0 + kV/2} +
\frac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{5} \\
\end{align}
The conversion, using Eq. (5), yields
\begin{align}
X_2 &= \dfrac{C_{A0} - C_{A3}}{C_{A0}} \\
X_2 &= \dfrac{C_{A0} - \dfrac{\alpha^2 v_0C_{A0}}{\alpha v_0 + kV/2} -
\dfrac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2}}{C_{A0}} \\
X_2 &= 1 - \frac{\alpha^2 v_0}{\alpha v_0 + kV/2} -
\frac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{6} \\
\end{align}
Now we combine Eq. (2) with Eq. (6)
\begin{align}
X_2 &= 1 - \frac{\alpha^2 v_0}{\alpha v_0 + \dfrac{X_1v_0}{2(1 - X_1)}} -
\frac{(1 - \alpha)^2v_0}{(1 - \alpha)v_0 + \dfrac{X_1v_0}{2(1 - X_1)}} \\
X_2 &= 1 - \frac{\alpha^2}{\alpha + \dfrac{X_1}{2(1 - X_1)}} -
\frac{(1 - \alpha)^2}{(1 - \alpha) + \dfrac{X_1}{2(1 - X_1)}} \tag{7} \\
\end{align}
Eq. (7) is a non-linear equation that we need to solve. Note that if $\alpha = 1/2$, the left-hand side yields 0.8, which is reasonable.
It is informed to us that $X_2 = 0.786$ and that the original conversion was $X_1 = 0.8$. Upon solving numerically, we found actually two possible solutions
$$ \boxed{\alpha \in (0.33500,0.66500)} $$
However, since both branches are "identical" to each other, it is the same if 33.5% goes up, and 66.5% goes down, or vice versa.
