I hope this is not redundant. The first time I tried to post it seems to have evaporated.
Before anyone calculates trip time with an "assumed" acceleration, some simple calculations should be done (and shame on the headline writer for not doing them).
The key number is specific power (W/kg).
F=ma (Force equals Mass times Acceleration so....
a = F/m
From a slightly old source, the "long term goal" for specific power of solar panels on spacecraft was 300 W/kg, and this was at Earth orbit, It will decrease on the way out. If we use the optimistic 300 and the optimistic Chinese data with .72 Newtons/2500 Watts we get:
a = (0.72 kg-meter)/(2500 Watt-second^2) * 300 Watt/kg
a = 0.086 m/s^2
1 g is 9.8 m/s^2 so we have less than 0.01 g (without any spacecraft body or payload).
There is a rule of thumb that the power source is 25-35% of a satellite's dry mass, which means we get (assuming the optimistic-for-payload smaller number) less than 0.0025 g acceleration.
Based on previous calculations by Pet Peeve, this gives us a 14 month straight line to Mars time and (if I read it correctly) that is straight acceleration without slowing down. This is an impressive number, but no big improvement over what we can do now.
It may be possible to better this if the craft takes a slingshot journey near the sun, since it will be able to greatly increase the acceleration while it's in close.
Don't look towards nuclear power to improve on this. The now cancelled ASRG was supposed to improve on present RTG (Radioisotope Thermoelectric Generator) technology and it would still have had a specific power of only 7 W/kg. This is useful out around Jupiter, where sunlight is dim, but not for the topic under discussion.