For a hexagon, wind loading should look like the area of one wall 90 degrees to the wind or perpendicular (24’ x 24’ for easy round number, this includes 2 above-water decks) or 576 sq ft Plus the two additional sides partially presented to the wind (1152 sq ft) and account for them being 60 degrees off, or presented 30 degrees to the wind. We’ll multiply the partially-presented side area by 1/3 to get 384 and add that to the perpendicular wall.
Our surface area presented to the wind is therefore effectively 960 sq ft.
Let’s postulate a fairly common ocean wind velocity of 30 mph. The simple formula for wind pressure P in imperial units (pounds per square foot) is P=0.00256*Velocity squared, where V is the speed of the wind in miles per hour (mph). We come up with 2.304 PSF * 960 = 2212 lbs of force. About a ton. Double the wind speed to 60 mph and the sideways force goes up to about 8847 lbs or 4.5 tons.
The above assumes that the wind is working 100% on one module and not at all on the other, which seems unlikely. But it does show a fairly high wind speed has a reasonable force applied to these modules, that could be counter-acted by a reasonably-sized weight (especially given you probably have a weight on each side for balance).