Fraction of concentration of benzene in air water
You don't need the original concentration or any of the variables you are missing. Henry's Law gives you everything you need. The usual formulation of Henry's Law at equilibrium:

$$c_{i,s}=k_{H,i,s}c_{i,s}$$

Where $c_{i,air}$ is the concentration of solute $i$ in the vapor above the solution, $k_{H,i,s}$ is the Henry's Law constant for solute $i$ in solvent $s$, and $c_{i,sol}$ is the concentration of $i$ remaining in solution.

We can rearrange Henry's Law to give you what you need - a ratio of the amount of benzene in the air to the amount in the water:

$$k_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}$$

The other snippet of information you need is the relationship between moles and concentration. There is a reason you were given volumes. 

$$n_{air}=c_{air}*V_{air}$$
$$n_{sol}=c_{sol}*V_{sol}$$

You only need the ratio of $\dfrac{n_{air}}{n_{sol}}$, so you really only have one unknown. It's algebra time!

$$K_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}=\dfrac{\dfrac{n_{air}}{V_{air}}}{\dfrac{n_{sol}}{V_{sol}}}$$

$$\dfrac{K_{H,i,s}V_{air}}{V_{sol}}=\dfrac{n_{air}}{n_{sol}}$$