Why will i calculate the wrong power ,but right loading resistance in this circuit?
You correctly found the Thevenin resistance of 3 ohm but then it went wrong when you assumed all the resistors were in series with 60 volts. The 3 ohms calculated is also the load you need to apply between A and B to get maximum power transfer too.


  Why can't the voltage of loading resistor be calculated as i
  think??where am i wrong?


Consider that this can be solved by "splitting the voltage source" covered in E5 is this short document: -



And that circuit becomes the one below when you convert the potential dividers on each supply and, calculate the new voltage source value for each side: -



So, the difference voltage between the two sources is 12 volts and it is this 12 volts that drives current through the 3 ohm resistor connected across A and B. So, that's: -

$$I = \dfrac{12\text{ volts}}{2.1 + 3 + 0.9} = 2 \text{ amps}$$

So the maximum power into the Thevenin resistance of 3 ohms is 12 watts.