The force on an electron due to electric field is given by $F=qE$. For a circularly polarized light, $E=E_x+iE_y$. So $F=q(E_x+iE_y)$ which means particle travels in a circle. As there is magnetic field associated with light, the particle experiences Lorentz force in the axial direction. This makes particle to travel in a helical path.
Please let me know if there is any mistake in my analysis.
No mistake, but if you are considering the Lorentz force on a charged particle due to an EM wave then consider this: The B-field amplitude is $c$ times less than the E-field amplitude. Therefore the magnitude of the force on the charged particle due to the B-field component of the Lorentz force ($=qvB$) is $v/c$ times less than the force due to the E-field component ($=qE$), because $E=cB$ for an EM wave in vacuum.
So unless you are making your electron move relativistically then the Lorentz force due to the B-field can be ignored. If the electron will be moving relativistically, then you'll need to solve a problem something along these lines.
[EDIT: There seems to be some confusion here: The E-field of circularly polarised light (e.g. travelling in the z-direction) could be written as $$ \vec{E} = E_0 \sin (\omega t - kz) \hat{i} + E_0 \cos(\omega t -kz) \hat{j}$$ Thus the acceleration of an electron at $z=0$ is given by $$\ddot{\vec{r}} = \frac{e}{m_e} E_0 \sin(\omega t)\hat{i} + \frac{e}{m_e} E_0 \cos(\omega t) \hat{j}$$ where the contribution of the B-field to the Lorentz force has been ignored for the reasons discussed above.
Integrating twice with respect to time and ignoring the constants of integration (assume the electron starts from rest at some coordinate with $z=0$ - or to put it another way, let's assume that $\omega$ is the frequency in the rest frame of the electron and that $\vec{r}$ is the displacement in that rest frame): $$\vec{r} = -\frac{e}{m_e \omega^2} \sin(\omega t) \hat{i} - \frac{e}{m_e \omega^2} \cos(\omega t) \hat{j},$$ which is circular motion in the $x,y$ plane of the electron's initial rest frame.]
