firstly this line wouldn't work:

> byte  b = 203;

as number literals are of type int  so you'd need to do this:

byte b = (byte)203;

After this line, b contains 11001011. When this line executes:

> int i = (int)b;

It is extended to 32 bits to make it int. The most significant bit is
1 so the rest is filled with 1s like this:

11111111111111111111111111001011

which is -53

When you execute  this line:

> int i = (int)b & 0xFF

You're doing this:

11111111111111111111111111001011 & 11111111

which just gives you 11001011 or 203.