Interpenetrating charged ball and wire.

Note: It is going to be a pretty long solution, so buckle up.

Firstly,
We find the force of interaction between the wire and the ball. To do that, we have 2 choices. Either we find the Field due to the wire at every point on the sphere, and then integrate the expression.....or a better choice is, to calculate the Field due to sphere at every point on the wire and then Integrate to find the force.

Now,
An important part is to split our calculations into 2 parts...First is the inside of the ball, and then second is the outside of the ball.

Note: For all calculations, I will consider $y$ to be the distance along the rod, measured from the symmetry point, and $dy$ to be the elemental length along $y$ .

1)For the First part, i.e inside of the ball,
Field at $y = \displaystyle\frac{kQ}{R^3} \sqrt{x^2+y^2}$
Due to symmetry reasons, we just take the vertically downward component of this field, i.e
$E = \displaystyle\frac{kQ}{R^3} \sqrt{x^2+y^2}\cdot\frac{x}{\sqrt{x^2+y^2}}$
$E = \displaystyle\frac{kQ}{R^3} x$
Hence,
Force on an elemental charge at $y = dq\cdot E$
$dF_1 = \displaystyle\lambda\cdot dy (\frac{kQ}{R^3} x)$
Since this is the First part,
$-\sqrt{R^2-x^2}\leq y \leq \sqrt{R^2-x^2}$
Hence,
Integrating on these limits,

$F_1 = \displaystyle\int\limits_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}\lambda\cdot (\frac{kQ}{R^3} x) dy$

$F_1 = \displaystyle\frac{2kQ\lambda x}{R^3} \sqrt{R^2-x^2}$

2)For the Second part, i.e outside of the ball,
Field at $y = \displaystyle\frac{kQ}{x^2+y^2}$ Due to symmetry reasons, we again take the vertically downward component of this field, i.e
$E = \displaystyle\frac{kQ}{x^2+y^2}\cdot\frac{x}{\sqrt{x^2+y^2}}$
$E = \displaystyle\frac{kQ}{(x^2+y^2)^{\frac{3}{2}}}$
Hence,
Force on an elemental charge at $y = dq\cdot E$
$dF_2 = \displaystyle\lambda\cdot dy\frac{kQ}{(x^2+y^2)^{\frac{3}{2}}}$
Since this is the Second part,
$-\infty<y\leq-\sqrt{R^2-x^2}$ and $\sqrt{R^2-x^2}\leq y<\infty$ Hence,
we will now integrate over one of these 2 limits, and double our answer once we are finished.

$F_2 = \displaystyle\int\limits_{\sqrt{R^2-x^2}}^{\infty}\lambda\cdot dy\frac{kQ}{(x^2+y^2)^{\frac{3}{2}}}$

$F_2 = \displaystyle\frac{2kQ\lambda}{x}(1-\frac{\sqrt{R^2-x^2}}{R})$

Now,
Finally we add these two Forces to get the Final force, as,
$F = F_1+F_2$

Note: Before writing out the big expression, we will use an approximation which I shall derive below.
For $x\to 0$ ,
$\displaystyle\sqrt{R^2-x^2} = R\sqrt{1-(\frac{x}{R})^2} \approx R(1-\frac{x^2}{2R^2})$

Note: With my Deepest apologies, I leave the simplification and the approximation of $F$ to the reader. I shall also request him/her to neglect the term with $x^3$ in it, as $x\to 0$

Therefore, we get,

$F = \displaystyle\frac{2kQ\lambda}{R^2} x$
Let $K = \displaystyle\frac{2kQ\lambda}{R^2}$
$\Rightarrow F = Kx$

Before we proceed , we shall use the concept of Reduced Mass , as this is an SHM. Again, I leave it to the reader to confirm and digest the concept of Reduced Mass before proceeding.

Reduced Mass $\displaystyle\mu = \frac{mM}{M+m}$

Hence,
Using $F = Kx$ ,
we have,

$\omega^2 = \displaystyle\frac{K}{\mu}$

$\omega = \displaystyle\sqrt{\frac{3kQ\lambda}{R^2} \frac{M+m}{mM}}$

As $\omega = 2\pi f$ , where $f$ is the frequency of oscillation,

$f = \displaystyle\frac{1}{2\pi}\sqrt{\frac{3kQ\lambda}{R^2} \frac{M+m}{mM}}$

$\Rightarrow f \approx \boxed{124Hz}$

That was quite a problem..It took me more than an hour to write this solution, so please be nice if you find some typos or any other conceptual mistakes..