Analog input to ATMEGA328P using op-amp - how does this circuit work?
Solving for V2: Vout=Vin * R2/(R1+R2)= Vin* 0.3299
So at 10V at V2 you get 3.3V output.



simulate this circuit &ndash; Schematic created using CircuitLab
Solving for I2:
Vout = I2*165
Vout = 20mA*165 = 3.3V



simulate this circuit
In order to use this circuit with 3.3V supply, you should use a rail to rail opamp in order to maximize the output range and then recalculate resistors, so that the output voltage would be in range of the max. output of opamp. For example, instead to have a voltage divider of 0.33 for V2, you use a divider of 0.3. At 10V input you get only 3V output, you loose that 0.3 voltage range of ADC.

You can see, that LM224D can output only Vcc-1.5V, so not suitable for you, if you don't supply it with at least 5V.
EDIT:
If you will use the 3.3V supply, then also a 2.2k output limiting resistor is not needed.
EDIT2:
Recalculate resistors for 0-3V output voltage:

\$R_{shunt}=\frac{3V}{20mA} = 150 \Omega\$, so the same R54, without/shorted  R32(15 Ohm).
From solving for V2. R2=47.15, R1=?

$$0.3= \frac{R_2}{R_1+R_2}$$
$$0.3\cdot(R_1+R_2)= R_2$$
$$0.3\cdot R_1+0.3\cdot R_2= R_2$$
$$0.3\cdot R_1= R_2\cdot (1-0.3)$$
$$ R_1= R_2\frac{0.7}{0.3}= R_2\cdot 2.333$$
$$ R_1= 110.01 k\Omega$$
$$ R_{50}= 110.01-2\cdot47= 16.01k\Omega \approx 16k\Omega$$