How to determine force constant of C-C bond in ethanol?
You can estimate the force constant (as I have done here) and calculate the frequency of the stretch or, as you request, go the other way. Regardless, you have 2 unknowns and thus the problem you present isn't solvable. Below, I work it from the assumed value of the force constant: to go the other way, you'll need to locate the absorption spectral line for the C-C stretch in ethanol and go from there. Either way, you'll need to make some assumptions regarding reduced mass.

From page 496 in Organic Chemistry by Brown, Iverson, Anslyn, and Foote, the formula

$$\tilde{\nu} = 4.12\;\sqrt{{k\over\mu}}$$

is given, where $\tilde{\nu}$ is the frequency in wavenumbers, $k$ is the force constant in units of dyne/cm, and $\mu$ is the reduced mass (in amu, or atomic mass units). They state that for a single bond, the force constant can be taken as $5\cdot 10^{5}$ dynes/cm (an approximation, and we do have to start somewhere). The 4.12 differs from the constant in your given equation due to unit conversion between dynes/cm and N/m.

Now, on to your problem. The reduced mass term is (generally)

$$\mu = {m_{1}m_{2}\over m_{1} + m_{2}}$$

which is for two masses. I posit that in ethanol, one carbon has a "mass" of 15 amu (that's the carbon with 3 hydrogens attached to it) and the other has a "mass" of 31 amu (the carbon plus 3 hydrogens and the oxygen). To go beyond this involves a significant amount of advanced work that is beyond the scope of your problem (but if you're interested, Molecular Vibrations by Wilson, Decius, and Cross is the place to start).

We calculate

$$\mu = {15\times 31\over 15+31} = 10.1$$

and substitute into the first equation

$$\tilde{\nu} = 4.12\;\sqrt{{5\cdot 10^{5}\over 10.1}} = 916\;\mathrm{cm}^{-1}$$

From some notes for a class at Caltech, it appears we can expect a carbon-carbon single bond to appear anywhere in the range of 800 to 1200 cm$^{-1}$. 

The calculated number for the C-C stretch here (towards the lower range we'd expect) is due to our "bloated" reduced masses, and that makes intuitive sense (if we do the same calculation for ethane, with "lighter" reduced masses for each carbon, we get $\tilde{\nu} = 1063$ cm$^{-1}$).