Calculating the 12V battery requirement for a 240V invertere system
You've forgotten that the battery is at 12 V.

Let's work it out in watts instead and then see what current we need at 12 V.


Warm up: 1200 W for one minute. We can really ignore this except for the piak current calculation.
Running: Since \$ P \propto V^2 \$ we can say that running power = \$ \frac {90^2}{240^2}1200 = 168 \ \text W \$. 
From \$ P = VI \$ we can calculate that \$ I = \frac {P}{V} = \frac {168}{12} = 14 \ \text A \$.
14 A for 6 hours = 84 Ah at 240 V.
Input current will be \$ \frac {output}{efficiency} \$ so at, say, 80% inverter efficiency you will require 17.5 A in and 105 Ah capacity.


Now let's look at the peak demand again:


1200 W at 12 V will require 100 A for a minute. When we factor in the inverter efficiency again we get 125 A peak current from the 12 V battery.


You probably require 2 x 125 Ah batteries. In that case go for a 24 V inverter and connect the batteries in series. The current and cable sizes will be lower.