detailed solution

Correct option is D

 
When the particle moves along a circle in the magnetic field B, the magnetic force is radially inward. If an electric field of proper magnitude is switched on which is directed radially outwards, the particle may experience no force. It will then move along  a straight line with uniform velocity. This will be the case when
$\mathrm{qE}=\mathrm{qvB}\text{ or }\mathrm{E}=\mathrm{vB}$
The radius of the circle in a magnetic field is given by $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$
Or, $\mathrm{v}=\frac{\mathrm{rqB}}{\mathrm{m}}$
$=\frac{\left(5.0\times {10}^{-2}\mathrm{m}\right)\left(20\times {10}^{-6}\mathrm{C}\right)(1.0\mathrm{T})}{20\times {10}^{-9}\mathrm{kg}}=50{\mathrm{ms}}^{-1}$

The required electric field is
=  $\mathrm{E}=\mathrm{vB}=\left(50{\mathrm{ms}}^{-1}\right)(1.0\mathrm{T})=50{\mathrm{Vm}}^{-1}$
This field will be in a direction which is radially outward at P