Here's how I did it for anyone who's interested.
From the density at STP I calculated the molecular weight 22.4x1.232 = 27.6
Given the info it was a gas comprised of only 2 elements limited the options for the calulated molecular weight to CO, C2
As far as I knew of those three only the diborane would react with water to give an acid.
That matched the question giving the acid as only one of the products of the reaction with water, hydrogen being the other.
A quick google showed the boric acid solubilty at 25°C was similar to the value given in the problem at 30°C.
Given 2 moles reacting would make 4 moles of boric acid and consume 12 moles of water it was just a case of setting up the equation to calculate the total mass of 4 moles of boric acid and how much water was left for it to dissolve in. Once I had the amount of boric acid in solution the rest must be the solid formed.
4 moles of boric acid weigh 247.3g
12 moles of water is 216.2g
The water left for the acid to dissolve in is 500-216.2 = 283.8g
Given 6.4g/100ml as the solubilty then 18.2g of the boric acid dissolved so (247.3-18.2) = 229.1g was undissolved solid.