Normally the, a burden resistor is chosen so that at the input primary side the impedance presented at the primary is in the micro-ohm to very low milli-ohm regionrange. Your CT steps down current by 20:1 and, with a 50 Ω resistor as burden, the burdenprimary impedance will "look like" 50 Ω ÷ 400 at the primary side. That's 80 milli-ohm and, almost certainly, the core of the CT will badly overheat and you'll get problems.
The coreIt will overheat because all the primary current will be flowing into the magnetization inductance of the primary. This might be 10 μΩ and, at 50 Hz the reactance will be about 3 milli-ohm3 milli-ohm reactive. Clearly the magnetization inductance takes the brunt of the primary current with very little energy being transferred to the secondary burden.
CTs are not expecting this to happen; they are not designed to take a high magnetization current and the core will saturate.
So, if you had 100 amps flowing, the primary (even though it looks like a short thick piece of wire) will develop a voltage of about 3 mill-ohm3 mill-ohm × 100 = 0.3 volts. This means the secondary will be 20 times higher at 6 volts. But, the main problem is that you should not run a CT with an inadequate burden else it will burn.
So, If I placed a 50 ohm resistor across seconder coil, Will I get 250Vac?
Not a chance.
Second question is that can I regulate this maximum voltage using multi-stage transistor-zener based voltage regulator to power op-amps, microcontroller, LCD display etc.
If you can prevent the CT from burning there is a small chance you might be able to get a regulated 3.3 volts by using a bridge and shunt regulator.