The output current required from the 3.3V
regulator is
1μA×0.999+50mA×0.001+25uA=75.999uA
The output power is
3.3V×75.999μA=250.8μW
The input power to the regulator is
250.8μW0.8=313.5μW
When the batteries are fully charged
the input current to the regulator is
313.5μW3V=104μA
If the batteries have a flat discharge
curve then you will get a life of
2700mAh104μA = 25837 hours = 2.95 years
Since your batteries are 2700mAh 1.5V AA
I am guessing that the discharge curve
is not flat. You will need to draw
higher currents as the voltage drops.
Also your regulator efficiency probably
drops at lower voltages. Again I am
guessing since I have not seen the
design.
Be careful when calculating using currents.
You may be inadvertently assuming that
the input and output voltage of the regulator
is the same. With an input of 3V and an output
of 3.3V it is not a big error. If you do
a more accurate estimate of the battery
discharge curve it will matter.