Calculations for current transformer and output voltage
$$\boxed{\text{Right up front - your burden resistor is far too high for this design}}$$
The whole point about CTs is that the primary magnetization inductive reactance is overwhelmed by the reflected burden resistance. Your turns ratio is 1:30 because that equals: -
$$\sqrt{\dfrac{\text{180 mH}}{\text{200 μH}}}$$
And the 600 Ω load reflects to the primary as a resistance of: -
$$\text{600 Ω}\cdot\dfrac{1}{900} = 0.667 \text{ Ω}$$
But, at 50 Hz the primary reactance is: -
$$2\pi\cdot 50\cdot 200\text{ μH} = 0.0628\text{ Ω}$$
And that impedance is certainly not overwhelmed by the reflected secondary impedance of 0.667 Ω. In other words, your burden resistor on the secondary needs to be a whole lot lower than 600 Ω. A whole lot lower if you want accuracy.

How do I calculate the value of the resistor on the output for a given
voltage?

You don't because you are constrained by other things.
So, you determine the primary magnetization inductive reactance (0.0628 Ω for 200 μH at 50 Hz) and you say to yourself: -
$$\boxed{\text{I need to make my burden reflected impedance much lower}}$$
So, if you went for something like 100 times lower i.e. 0.0006 Ω you would find that when transferred to the secondary, it is a resistance of: -
$$\text{0.0006 Ω}\times 900 = 0.54 Ω$$
That would be a good value for a burden resistor.

Secondary resistance as stated by the OP: -
Resistance = 12 ohms

This is too much for your design. Can you see that based on the above calculations? It's the same story with your primary resistance of 0.2 Ω. You just cannot expect a CT to operate anything like a CT is meant to operate with these resistance values. Just think about it; you say the primary current is 10 amps and that means your primary winding is dissipating 20 watts of \$I^2R\$ power losses. Be realistic.