Don't hammer the charge particle , he is innocent !

First we need to evaluate the electric field by a cylinder of $\color{#D61F06}{ \text{uniform} }$ charge density $\rho$ , length $L$ and radius $R$ at a point on its axis . This can be found by integrating over elemental discs . I will leave the integration part and and only give the final result here .

$\displaystyle E = \dfrac{ \rho ( L - \sqrt{ (d+ L)^2 + R^2} + \sqrt{d^2 + R^2})}{2 \epsilon}$ ........ where $d$ is the distance of the point from the base of the cylinder closer to it

Consider the state of the particle when it is displaced by a small distance $x$ from its $\color{#D61F06}{ \text{ mean } }$ position

The $\color{#3D99F6}{ \text{net}}$ electric field acting on the particle is

$\displaystyle E = \frac{ \rho \Big( \big( \sqrt{( \frac{d}{2} + x + L )^2 + R^2 } - \sqrt{ ( \frac{d}{2} -x + L )^2 + R^2 } \big ) -\big( \sqrt{ (\frac{d}{2} + x)^2 + R^2} - \sqrt{( \frac{d}{2} -x )^2 + R^2 } \big) \Big )}{ 2 \epsilon}$

grouping the terms as shown above and after substituting the values given and simplifying using the $\color{#3D99F6}{ \text{binomial approximation} }$ $( (1 + x)^n \approx 1 + nx ,\text{ for} ~ x<< 1)$ , we get

$E = \dfrac{ \rho x ( 4\sqrt{5} - 5\sqrt{2} )}{10 \epsilon }$

the acceleration of the charged particle will be

$a = \dfrac{qE}{m}$

Comparing with standard equation for $\color{#3D99F6}{S.H.M}$ $a= \omega^2 x$ , we get

$\displaystyle \omega = \sqrt{ \frac{ \rho q ( 4\sqrt{5} - 5\sqrt{2} )}{10m\epsilon}}$

Thus time period of oscillation is

$\displaystyle \color{#3D99F6}{ T = 2 \pi \sqrt{ \frac{ (4\sqrt{5} + 5\sqrt{2}) m\epsilon }{ 3 \rho q} }}$

$\therefore \color{#D61F06}{a =4 , b=5 , c=5, d= 2, e =3 }$

Hence the answer is $\boxed {\color{#20A900}{19}}$