= Solution
Let
$$
g'_H=\frac{g(\rho_1-\rho_H)}{\rho_0},
\qquad
g'_L=\frac{g(\rho_1-\rho_L)}{\rho_0}.
$$
The global steady heat, or <buoyancy flux>, balance equates the floor-source input to the buoyancy carried out by the one-way upper-layer exhaust:
$$
B=Q_Vg'_H.
$$
The corresponding upper-to-lower density jump is fixed by the warm plume crossing the interface,
$$
B=Q_B(h)(g'_H-g'_L).
$$
The first balance also shows that the descending cold plume has buoyancy-flux magnitude per unit wall length
$$
\mathcal B=\frac{Q_Vg'_H}{L}=\frac BL.
$$
At a steady interface, the upward axisymmetric-plume volume flux equals the total downward wall-plume volume flux. Using parts (c) and (d),
$$
C_PB^{1/3}h^{5/3}
=L\alpha(z_o-h)
\left(\frac{B}{\alpha L}\right)^{1/3}.
$$
After cancellation of $B^{1/3}$, the required implicit geometric relation is
$$
\boxed{
C_Ph^{5/3}
=\alpha^{2/3}L^{2/3}(z_o-h),
\qquad
z_o=z_V+\frac{H'}{2\alpha}
}.
$$
Thus the ideal steady interface fraction is independent of the source strength: increasing $B$ multiplies both opposing plume volume fluxes by $B^{1/3}$. The floor area $A$ and room height $H$ affect the transient filling time and admissibility of the assumed ordering, but not this steady integral balance, provided $0<h<z_V$ and the plumes remain separated.
Set
$$
K_V=\frac{C_d}{3}LH'^{3/2}.
$$
Combining the <single-opening exchange flow> relation $Q_V=K_V\sqrt{g'_H}$ with $B=Q_Vg'_H$ gives
$$
\boxed{
Q_V=(K_V^2B)^{1/3}
=\left(\frac{C_d^2L^2H'^3B}{9}\right)^{1/3}
}.
$$
It also gives $g'_H=(B/K_V)^{2/3}$, after which the second density balance determines $g'_L$.
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