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.
Realistically, ten minutes is about all you could probably hope for.
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.
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