Travelling at a speed of 2.0 × 10^5 m s^ -1 , a proton passes...

When the proton moves undeflected in crossed electric and magnetic fields, the electric force balances the magnetic force:

$qE = qvB \Rightarrow E = vB$

When the electric field is switched off, the proton moves in a circular path under the influence of the magnetic field. The radius of the circular path is given by:

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

Rearranging for the magnetic field $B$:

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

Substituting the given values ($m = 1.6 \times 10^{-27} \text{ kg}$, $v = 2.0 \times 10^5 \text{ m s}^{-1}$, $q = 1.6 \times 10^{-19} \text{ C}$, $r = 4.0 \times 10^{-2} \text{ m}$):

$B = \dfrac{1.6 \times 10^{-27} \times 2.0 \times 10^5}{1.6 \times 10^{-19} \times 4.0 \times 10^{-2}}$

$B = \dfrac{2.0 \times 10^{-22}}{4.0 \times 10^{-21}} = 0.05 \text{ T}$

Now, substituting $B$ to find the electric field $E$:

$E = vB = (2.0 \times 10^5) \times 0.05$

$E = 1.0 \times 10^4 \text{ N C}^{-1}$