= Solution
Take an explicitly illustrative European annual mean ground-level irradiance of $100\,\mathrm{W\,m^{-2}}$, the stated conversion fraction $0.005$, and dry carbohydrate heating value $16\,\mathrm{MJ\,kg^{-1}}$. One square kilometre then produces
$$
M=\frac{(100)(10^6)(3.156\times10^7)(0.005)}{16\times10^6}
=9.86\times10^5\,\mathrm{kg\,year^{-1}}.
$$
Hence $\boxed{M\sim10^3\,\mathrm{tonnes\,km^{-2}\,year^{-1}}}$, or about $10\,\mathrm{tonnes\,ha^{-1}\,year^{-1}}$, as a dry-mass order-of-magnitude estimate. The assumed irradiance is not a site-specific value supplied by the paper. The result scales linearly with irradiance and conversion efficiency and inversely with heating value; wet harvested mass would additionally depend on water content.
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