I don't have a good means of measuring capacity, but two of them weigh 80g.
Might be a good battery, more than twice as heavy as my batteries.
For a rough capacity estimation you could try it with a resistor, for example 10 ohm (needs to be a 2 W type, use multiple in series and parallel, or a higher value, which then needs longer). If the capacity is not that high, this will drain it in a few hours max, so keep an eye on it that you don't deep discharge it and stop discharging it when it reaches 3 V, this should be safe for most 18650 accus. Then use the middle of the discharge curve, 3.7 V might be a good approximation. I guess this is what the "rated" voltage means on the battery. This means it delivers on average 3.7 V * 0.37 A = 1.4 W. If it can do this for one hour, the energy stored in the battery is 1.4 Wh. If only for half an hour, then 0.7 Wh etc. You get the non-SI value "mAh", if you divide the energy by the rated voltage, so 0.7 Wh / 3.7 V = 0.19 Ah = 190 mAh. There is even a webpage for it:
https://milliamps-watts.appspot.comOr measure the current your flashlight needs and measure the time until it is dim. But could be tricky, if it uses a switched-mode power supply.
BTW, I just verified my electronic load with a power supply, and it uses always 1 V as the rated voltage for battery capacity measurement. Misunderstandings like this happens, when you use silly units like AH. Couldn't find anything in the manual how to change it. So the reading "0.4 AH" means the battery really has 0.4 * 3.7 = 1480 mAh (or maybe a bit less, if the average voltage is lower), because it assumed 1 V, and not 400 mAh as the display suggested. Still about 7 times less than what was sold.
So a recommendation: don't buy the RK8511 electronic load. The capacity measurement is useless, unless I missed something how to change the unit, because it should really measure Wh. I verified it with a power supply to be sure: it shows the same value, 0.0168 AH, when using 9.93 V with 1 A for a minute, and 4.939 V with 1 A for a minute, it just ignores the input voltage, and the entered cut-off voltage. Useless. I guess this is why the seller got suspicious, it can't be
that bad