What happens when a semipermeable membrane allows only one solute to pass through?
I am afraid you are stuck because you considered glucose as osmotically active. 
But you assumed that glucose can freely move across the membrane, which is the opposite of osmotic activity. 
So, short-short answer : ignore the glucose, it is an osmolyte, but it is not osmotically active on the membrane you are thinking of; its concentration is going to equilibrate on each side of the membrane :
$$[\mathrm{C_6H_{12}O_6}] = \frac{1 \mathrm{mol}}{2 \mathrm{L}} = 0.5\, \mathrm{mol\, L^{-1}}$$
This solution of glucose is simply the solvent. 
Also, you may have inverted terms in the equilibrium equation.

A longer answer : 
First we assume that the volume and density of the solution are unchanged by the solutes added (ideal solutions).
We are left with $1 \,\mathrm{osmol\cdot L^{-1}}$ NaCl on one side of the membrane, and no osmolyte on the other side. 

The equilibrium between hydrostatic and osmotic pressure ($i c R T $) then writes : 
$$ 2 \rho g y = i \frac{n}{(h+y)A} RT  $$
Solving for $y$, we get 
$$ 2y = \sqrt{\frac{2inRT}{\rho g A}+h^2}-h$$
Using the following values :  $i=2$, $n=0.5 \,\mathrm{mol}$, $R=8.3\,\mathrm{kg\cdot  m^2\cdot s^{-2}\cdot K^{-1}\cdot mol^{-1}} $, $T=3\cdot 10^2\,\mathrm{K}$, $\rho=10^3\,\mathrm{kg\cdot m^{-3}}$, $g=9.8\,\mathrm{m\cdot s^{-2}}$ ,  $A=10^{-2}\, \mathrm{m^{-2}} $ and $h = 10^{-1} \mathrm{m} $, we obtain :
$$ 2y \simeq 7\, \mathrm{m} $$
This value is actually unreachable given the volume of solvent and the geometry of your beaker. All the glucosed water would be drawn on one side, and equilibrium could not be reached.

The osmotic pressure of a solution containing $1\,\mathrm{osmol\cdot L^{-1}}=10^3 \,\mathrm{osmol\cdot m^{-3}} $ is approximately $\Pi = 2.5\, \mathrm{MPa} = 25 \, \mathrm{bar}$. To equilibrate such a pressure with hydrostatic pressure, you need a column of water of depth $z = \frac{\Pi}{\rho g} \simeq 260\,\mathrm{m}$. 

The osmolarity of sea water is about $1\,\mathrm{osmol\cdot L^{-1}}$. In a desalination plant, sea water is purified by reverse osmosis using pressures from 50 to 80 bar. Osmotic equilibrium is of utmost importance in physiology, but this you must know better than me. Unless there is active transport, diffusion of solutes, or extraordinary high applied pressures (which may be relevant for deep-sea creatures), water is always going from the least to the most concentrated side of the membranes, so as to even out the osmolarities. Hope it helped.