How much water vapour can a gram of Uppsalite absorb at NTP?
One fallacy in your question is assuming that the particles 1) are spherical and 2) only accessible on the outside. The press release linked says that this material is nanoporous - the particles are perforated by pores "all having a diameter smaller than 10 nano meters". The extra surface area is inside these pores. Spheres have the lowest surface-area-to-volume ratio. Other shapes have higher ratios (in other words, more surface area for less volume). These nanoporous materials maximize the surface area to volume ration - minimizing the volume. If these particles are $\ce{MgCO3}$, then their density is $\pu{2.958 g/cm^3}$ (Wikipedia). One gram of this material has a volume of:

$$\pu{1 g}\times \pu{1 cm^3 g^-1}\times \left(\frac{\pu{1 m}}{\pu{10^2 cm}}\right)^2=\pu{3.38E-7 m^3}=\pu{3.38E-3 L}$$

The surface-area-to-volume ratio is then:
$$\frac{A}{V}=\dfrac{\pu{800 m^2}}{\pu{3.38E-7 m^3}}=\pu{2.37E9 m^-1}$$

A second fallacy is that the volume of water that can be absorbed is the same as the volume of the particle. The particle is not absorbing water per se. The water is not taken into the interior volume where the $\ce{MgCO3}$ is. That would require displacement of the magnesium carbonate, and where would it go? The water goes into the pore of the material - it adsorbs to the surface inside those pores. 

You can read more about the science at the original research article:  A Template-Free, Ultra-Adsorbing, High Surface Area Carbonate Nanostructure. PLoS ONE 2013, 8 (7). DOI: 10.1371/journal.pone.0068486.