Let the boat's speed in still water be \(b\) km/h and the current's speed be \(s\) km/h. From the given data, the two rates work out to \(50/5 = 10\) km/h and \(49/7 = 7\) km/h, corresponding to the boat's two directional speeds against and with the current.
Adding these two rates together and dividing by 2 gives the boat's speed in still water: \(b = (10+7)/2 = 8.5\) km/h, matching option 3.
Worth flagging: the numbers as given actually make the "upstream" rate (10 km/h) faster than the "downstream" rate (7 km/h), which is the reverse of what should physically happen, so the upstream/downstream labels in the source question may be swapped; the final still-water speed of 8.5 km/h works out the same either way since it's just the average of the two rates.