Rise and falls times are limited by the maximum slew rate the opamp can produce. This is something you look up in the datasheet. Note that the slew rate might be dependent on supply voltage and possibly other factors. Or, the datasheet might just give a single minimum large-signal slew rate for the whole range of possible conditions.
Once you have the slew rate, divide the voltage that has to be traversed for whatever you consider "rise" and "fall" by the slew rate. For example, if you decide the opamp can do 3 V/µs at minimum and you want the output to change 10 V, then that will take at most:
(10 V) / (3 V/µs) = 3.3 µs
The slew rate is 70v/ųs max.
I opened the datasheet from the link you provided, and right on the second line of the first page it says that only 50 V/µs is "ensured". The only mention of 70 V/µs I can see is for a typical value, which is meaningless.
(18 V) / (70V/µs) = 0.257143 ųs = 257.143 ns
That is correct, sort of. As stated before 70 V/µs is not the spec for this opamp. The other problem is that 257.143 ns is absurd. Carrying an extra digit or two in intermediate calculations makes sense to avoid adding additional noise. But claiming to know the transition time to within a picosecond is ridiculous. Just 257 ns is already lots more accurate than you know the actual slew rate, but would be an acceptable answer with the understanding this is an intermediate value with a bit more precision than really needed.
And then i did this: (257.143/5)×4 = 205.714 ns.
Because rise/fall time start at 10% and end at 90%.
Yes, you're really asking about slewing 80% of the 18 V, so the resulting time will be 80% of the time calculated for 18 V. Other than the absurd precision, this is correct for your starting assumptions.