This circuit is working, but Is there any way I can use A FET instead of those BJT's?

There are lots of simple circuits on the internet. e.g.

 

There are lots of simple circuits on the internet. e.g.

View attachment 204158

This simple circuit like others on the internet slowly destroys the BJT transistors because the maximum allowed reverse emitter-base voltage is exceeded (avalanche breakdown) each time a transistor turns off. The maximum allowed reverse emitter base voltage of most little transistors is only 5V or 6V but here they get almost 12V. A diode and resistor added to each transistor will fix it.

Google shows many circuits that have a multivibrator made with two Mosfets that produce a squarewave:

 

Is a square-wave out okay or would you prefer a sine-wave?
Of course that would require much more circuitry to generate a sine-wave modulated PMW signal to drive the transformer.

 

I have no Center tap transformer...

 

Is a square-wave out okay or would you prefer a sine-wave?
Of course that would require much more circuitry to generate a sine-wave modulated PMW signal to drive the transformer.

square wave without CT transformer will be ok

 

One way to get 120V peak out of that transformer is to drive the 12V side with a MOSFET bridge circuit from the 12V.
That will put ±12V across that 12V winding which will give ±120V from the 120V winding
You will need a proper bridge driver with non-overlapping signals to drive the bridge and avoid large current shoot-through during the switching time.
I have a circuit for that, if interested.

 

One way to get 120V peak out of that transformer is to drive the 12V side with a MOSFET bridge circuit from the 12V.
That will put ±12V across that 12V winding which will give ±120V from the 120V winding
You will need a proper bridge driver with non-overlapping signals to drive the bridge and avoid large current shoot-through during the switching time.
I have a circuit for that, if interested.

May I see please?

 

and oh, i f I add a diode between pin 7&2 of IC555,will I get 50%duty cycle?

 

and oh, i f I add a diode between pin 7&2 of IC555,will I get 50%duty cycle?

No, you need two diodes like this:

 

May I see please?

Below is the LTspice simulation of a non-overlapping clock driver with a complementary MOSFET bridge. The driver circuit uses three IC packages.
The NOR gates generate the non-overlapping function so that the top and bottom adjacent MOSFETs are never ON at the same time (the second and third magnified plots show this non-overlap for both rising and falling waveforms).
[Remember that the N-MOSFET gate (yellow) is high when ON, but the P-MOSFET gate (blue) is low when ON].
This prevents the high shoot-through current that would otherwise occur when the two transistors are momentarily on simultaneously during switching.

The red trace shows the ±12V waveform across the transformer winding (R_Trans).

Two CD4049 CMOS HEX inverter chips are used with three parallel outputs for each gate as a poor-man's push-pull circuit to drive the large MOSFET gate capacitance.

For driving a transformer, the input waveform should have an exact 50% duty-cycle.
That can be achieved by setting the 555 frequency to 100Hz and running that through a CD4013 flip-flop configured as a divide-by-2 toggle (/Q output to D input).
The 555 duty-cycle is thus not a factor, as the FF triggers on only one edge of the clock.

Note: Not shown, but required, are 100nF ceramic caps across each chip from their power pin to ground pin, as well a large (e.g. 100μF) cap from the bridge Vdd supply node to the bridge ground.