1. Condition for undeflected motion in crossed fields

For a charged particle to move undeflected in crossed electric and magnetic fields,

$qE = qvB$

So,

$E = vB$

2. When electric field is switched off

Now only the magnetic field acts, so the proton moves in a circular path.

Magnetic force provides the centripetal force:

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

Cancelling $v$:

$qB = \frac{mv}{r}$

Hence,

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

3. Substitute given values

Given:

$m = 1.6 \times 10^{-27}\,\text{kg}$ $v = 2 \times 10^5\,\text{m/s}$ $q = 1.6 \times 10^{-19}\,\text{C}$ $r = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m}$

So,

$B = \frac{(1.6 \times 10^{-27})(2 \times 10^5)}{(1.6 \times 10^{-19})(2 \times 10^{-2})}$

$B = \frac{3.2 \times 10^{-22}}{3.2 \times 10^{-21}} = 10^{-1}\,\text{T}$

Thus,

$B = 0.1\,\text{T}$

4. Now calculate electric field

Using

$E = vB$

$E = (2 \times 10^5)(0.1) = 2 \times 10^4\,\text{N/C}$

Given that

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

we get

$x = 2$

5. Comparison with stored answer

Derived answer: $2$

Stored correct answer: $2$

They match.