Normally, a burden resistor is chosen so that the impedance presented at the primary is in the micro-ohm to low milli-ohm range. Your CT steps down current by 20:1 and, with a 50 Ω burden, the primary impedance will "look like" 50 Ω ÷ 400. That's 80 milli-ohm and, almost certainly, the core of the CT will badly overheat.
It 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-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-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.
