Solution

a) Equality of forces in the equilibrium position

$$mg=\frac{kq^2}{R^2}\Rightarrow kq^2=mgR^2$$

$$F'=F-mg=\frac{mgR^2}{R^2+2Rx+x^2}-mg$$

$$F'=\frac{mgR^2-mgR^2-2mgRx-mgx^2}{(R+x)^2}$$

Using the approximation for $x\ll R$

$$F'\approx-\frac{2mgRx}{(R+x)^2}$$

Next we will use approximations: $(1+x)^\alpha

$$\approx1+\alpha x$, где $x\ll1$$$

F'=-\frac{2mgRx}{(R+x)^2}\approx-2mgRx\frac{1}{R^2}(1-2\frac{x}{R})

$$

$$

F'=-\frac{2mgx}{R}-\frac{4mgx^2}{R^2}

$$Given $x \ll R$, we neglect the term with $R^2$$$

\boxed{F'\approx-\frac{2mgx}{R}}

$$b) Next, we write down the system of equilibrium equations for the system$$

\left\{\begin{matrix}3mg=\frac{kq^2}{x^2} \\ mg=\frac{kq^2}{R^2} \end{matrix}\right.

$$Now let's find at what distance from the lower end of the spoke the equilibrium position of the body is located.$$

\frac{R}{x}=\sqrt{3}\Rightarrow \boxed{x=\frac{R}{\sqrt{3}}}

$$The force acting on a body depends on its small displacement from the equilibrium position$$

\boxed{F''=-\frac{2m'gx}{R'}=-\frac{6mgx}{R'}}$$