To calculated the power dissipation us Ohm's law.
P = VI
P = I2R
P = V2/R
In your case, all of the load current flows through the current sense resistor, which also has 1.25V across it, so use whichever formula is easiest for you.
The circuit requires plenty of voltage headroom. Ideally the input voltage should be at least 4V above the load voltage, but as you're only using it for 200mA, you'll probably get away with 3V, going from the graphs on the data sheet.