How to select an LDO voltage regulator?
You have understood both the dropout voltage and ground current wrong.

Dropout voltage is the difference between input and output that is at least needed for the regulator to work. It means that when your load is drawing 200mA, worst case you need only 235 mV more on the input. So as the output is 3.3V, you must have 3.535V or more at the input.

Ground current is indeed the difference between input and output current. But the regulator does not take in constant 250mA, so the difference is not 50mA. The datasheet says than when the load is drawing 200mA on output, the regulator itself takes only 6uA of current to work.

Based on those, you can use this. However you have not considered power dissipation yet. 5V in, 3.3V out, 0.2A current means the device must dissipate 0.34W. Depending on which package device you use, it can heat up by 91 degrees C, which means the ambient temperature must never exceed about 34 degrees C.