$\rm Lux$ and $W/m^2$ relationship?

There is no simple conversion, it depends on the wavelength or color of the light.

However, for the sun there is an approximate conversion of $0.0079 \, \text{W/m}^2$ per Lux.

To plug in numbers as an example: if we read 75,000 Lux on a light sensor, we convert that reading to $\text{W/m}^2$ as follows: $$75,000 \times 0.0079 = 590 \, \text{W/m}^2 \, .$$

Source

I'm afraid that it is not easily possible to take the luminous flux and obtain the insolation (as radiant flux). Here's why:

The luminous flux $F$ is calculated from the radiant spectral power distribution $J(\lambda)$ by weighting each wavelength with a luminosity function $y(\lambda)$ as per

$$ F = c \int J(\lambda)y(\lambda)\mathrm{d}\lambda$$

where $c$ is some unit conversion constant between lumen and watts. The total radiant flux $\Phi$ would be

$$ \Phi = \int J(\lambda)\mathrm{d}\lambda $$

The problem is that the calculation of the luminous flux is not invertible - portions of $J(\lambda)$ lying outside the visible range are cut off by the luminosity function being zero there, and it is perfectly possible for two $J(\lambda)$ of different radiant flux $\Phi$ to have a similar luminous flux $F$.

However, in the case of solar radiation, there might be a way - we know the spectral composition of sunlight, and so we know the form of $J(\lambda)$ already quite well - you could try to run an algorithm that fits the scaling of the known spectrum $J(\lambda)$ to yield the value of $F$ you measure and then calculate $\Phi$ from that spectral function. I'm not sure how good this idea is though.