Calibrate phase with samples shift

You should not be looking at A resistor, but several in series. It is easier to get a lower rated voltage resistors, and it one goes short-circuit (however unlikely that might be) you still have others in series to limit the current.

Where does your value of 65 kΩ come from? According to my sources, your mains voltage is 220 V, 50 Hz. The rated current of the c.t. is 2 mA, so at 220 V + 10%, the series resistance would be 121 kῺ. I would use 4 × 33 kΩ in series. At your nominal voltage + 10%, each would have approximately 60 V across it, the power in each would be about 0.11 W. For safety and long life, I would use 0.5 W resistors.

Again at nominal voltage+10%, the current is 1.833 mA, which is also the secondary current. therefore your burden will be (using my magic number 1.1 V for the rms voltage taking component tolerances into account) exactly 600 Ω. Even using a 560 Ω resistor (giving you 2.9 V peak-peak), will not degrade the performance of the voltage input, because unlike current, the range over which the (rms) voltage changes is minuscule when compared to the current.

You must still remember that the 4 resistors, their interconnections and the primary winding of the c.t. and its connections are all LIVE to mains voltage, and must be shrouded to prevent contact.