Gencis,
Before answering the question, let's have a look at your loads.
If I understand you correctly, you would like to use a 12v battery to power a 220vac inverter which will drive a 1800W steady-state load.
Your inverter is rated for 1500W but provides a 3000W short-term peak power which, for example, would allow for starting up a motor, but not for continuously driving anything at over 1500W. Thus, you have a problem as your inverter will be overloaded unless you find some way of reducing the power draw. For example, I have an 1800W induction cooktop that works off a sine-wave inverter just fine as long as I don't run it at full power.
Now, to your question -
For an output of 1800W if the inverter were 100% efficient and we run at a nominal 12v that would translate into 150A.
Inverter efficiency is unknown, but let's assume 80%. Thus your input 12v current draw is 187.5A.
Your battery capacity is 142Ah; however, battery capacity in ampere-hours is usually a 20-hour rating which means, in your case, 142Ah/20 = 7.1A can be drawn for 20 hours until the battery voltage drops to 10.5vdc.
Batteries are notoriously non-linear when it comes to available capacity vs. current draw (i.e., output power). The more current you draw, the lower the Ah rating.
For example, a very high-quality Deka Dominator Gel deep-cycle 4D battery has a 20-hr rating of 183Ah but a one-hour rating of 122Ah. In other words, one can draw 122Amps for only one hour from this 183Ah battery.
For argument's sake, let's use this ratio with your 142Ah battery: 142*122/183 = 95Ah is your battery's one-hour capacity; in other words, you should be able to draw 95A for an hour if it's a similar high-quality Gel battery.
But wait, you want to draw 187.5Amps! I have not found a capacity specification for less than one hour and the capacity continues to be non linear the more current you draw. If it were linear, you could assume roughly one-half hour.
At a full 1800W load, if the inverter works, I think you would be hard-pressed to get more than 15-20 minutes out of this battery before its voltage dropped to 10.5vdc, by which time the inverter would probably have turned itself off anyway.
Conclusion:
Realistically, ten minutes is about all you could probably hope for.
Edit
Addendum: the battery's CCA rating depends on whether it is SAEJ537, IEC, or DIN. Those tests are for very short-term starting-motor cold cranking capacity and don't apply to your inverter question. For example, the DIN CCA test takes your 850CCA battery frozen to -18°C and discharges it at 850A and to pass the test the voltage must stay above 9v for 30 seconds and 6v for 150 seconds.