Using that ideal white source on the wiki link, that states 251 lm/W.
The insolation level is $2.61kWh/m^2d = \frac{2610}{24}W/m^2$
which across $65cm^2 (= 0.0065m^2)$ gives 0.7W. If everything were 100% efficient, then you'd have $0.7 \times 251 lm = 178 lm$
Now we derate on some maximum theoretical efficiencies. The biggest theoretical derating will be on the PV, and that will completely dwarf any loss on the maximum theoretical efficiency of a round trip into storage and back.
For a single-junction n-p PV cell in unconcentrated sunlight, the maximum efficiency (Shockley–Queisser, DOI:10.1063/1.1736034) is 30%, giving you about 53 lm. By layering multiple junctions, you could theoretically get 42% (2 junctions, 74 lm), 49% (3 junctions, 87 lm), tending to 68% ($n\to\inf$, 121 lm) (doi: 10.1088/0022-3727/13/5/018)
Now, if you are allowed to put a concentrating lens onto the $65 cm^2$ cell, so that the sun jar harvests light from a much larger area, then we can really go to town. From the second link, the maximum efficiencies for concentrating PV, as the number of junctions tends to infinity, is 86.8%. So then you've got to find out what the maximum concentration could be without the whole lot bursting into flames ... that 86.8% is based on a concentration factor of 45 900 , so it's a PV cell made of pure unobtainium, and I'll stop right there.