Let's break down the problem step by step:

When the proton moves undeflected in crossed electric and magnetic fields, the electric and magnetic forces balance each other. That is,

$$qE = qvB,$$

which simplifies to

$$E = vB.$$

After the electric field is switched off, the proton moves in a circular path under the action of the magnetic force. The magnetic force provides the required centripetal force:

$$qvB = \frac{mv^2}{r}.$$

Solving for the magnetic field $$B$$, we get:

$$B = \frac{mv}{qr}.$$

Now substitute this expression for $$B$$ back into the equilibrium condition:

$$E = vB = v\left(\frac{mv}{qr}\right) = \frac{mv^2}{qr}.$$

Plug in the given values:

Proton mass, $$m = 1.6 \times 10^{-27} \, \text{kg}$$

Speed, $$v = 2 \times 10^5 \, \text{m/s}$$

Radius, $$r = 2 \, \text{cm} = 0.02 \, \text{m}$$

Proton charge, $$q = 1.6 \times 10^{-19} \, \text{C}$$

Thus,

$$E = \frac{(1.6 \times 10^{-27} \, \text{kg})(2 \times 10^5 \, \text{m/s})^2}{(1.6 \times 10^{-19} \, \text{C})(0.02 \, \text{m})}.$$

Calculate the numerator:

$$(2 \times 10^5)^2 = 4 \times 10^{10},$$

So, $$(1.6 \times 10^{-27}) \times (4 \times 10^{10}) = 6.4 \times 10^{-17}.$$

Calculate the denominator:

$$(1.6 \times 10^{-19}) \times (0.02) = 3.2 \times 10^{-21}.$$

Now compute the electric field:

$$E = \frac{6.4 \times 10^{-17}}{3.2 \times 10^{-21}} = 2 \times 10^4 \, \text{N/C}.$$

The problem states that the magnitude of the electric field is $$x \times 10^4 \, \text{N/C}.$$ Since we found

$$E = 2 \times 10^4 \, \text{N/C},$$

it follows that

$$x = 2.$$