Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The net force exerted on the middle charge due to the adjacent ions is the sum of the forces exerted by the two other charges from the left and the right.
The new distances between the middle charge and the two other charges, after displacing the middle charge a distance of $x$ where $x\ll b$, are $ (b + x) $ and $ (b - x) $.
Thus, the net force exerted on the middle charge is given by
$$
F_{\text{net}} =F_{\rm left}-F_{\rm right}= \frac{e^2}{4 \pi \epsilon_0 (b + x)^2} - \frac{e^2}{4 \pi \epsilon_0 (b - x)^2}
$$
$$
F_{\text{net}} = \frac{e^2}{4 \pi \epsilon_0 b^2\left(1+ \frac{x}{b}\right)^2} - \frac{e^2}{4 \pi \epsilon_0b^2 \left(1- \frac{x}{b}\right)^2}
$$
$$
F_{\text{net}} = \frac{e^2}{4 \pi \epsilon_0 b^2}\left(1+ \frac{x}{b}\right)^{-2} - \frac{e^2}{4 \pi \epsilon_0b^2} \left(1- \frac{x}{b}\right)^{-2}
$$
$$
F_{\text{net}} = \frac{e^2}{4 \pi \epsilon_0 b^2}\left[\left(1+ \frac{x}{b}\right)^{-2} - \left(1- \frac{x}{b}\right)^{-2}\right]\tag 1
$$
Since $ x \ll b $, we can use the binomial approximation to simplify.
So we need to apply the binomial approximation to
$$
\left[\left(1+ \frac{x}{b}\right)^{-2} - \left(1- \frac{x}{b}\right)^{-2}\right]\tag 2
$$
we need to use the binomial expansion for terms of the form $ (1 + u)^n $ when $ u $ is small. Specifically, for small values of $ \frac{x}{b} $, we can expand the terms in powers of $ \frac{x}{b} $.
We know for $ u\ll 1 $, the binomial expansion for $ (1 + u)^n $ is given by $ (1 + u)^n \approx 1 + n u + \frac{n(n-1)}{2} u^2 + \dots $
Now let's apply this to each term in the expression.
$$
\left(1+ \frac{x}{b}\right)^{-2} \approx 1 - 2 \left( \frac{x}{b} \right) + 3 \left( \frac{x}{b} \right)^2\tag 3
$$
And
$$
\left(1- \frac{x}{b}\right)^{-2} \approx 1 + 2 \left( \frac{x}{b} \right) + 3 \left( \frac{x}{b} \right)^2\tag 4
$$
Plug (3) and (4) into (2);
$$
\left[\left(1+ \frac{x}{b}\right)^{-2} - \left(1- \frac{x}{b}\right)^{-2}\right] \approx \left( 1 - 2 \frac{x}{b} + 3 \frac{x^2}{b^2} \right) - \left( 1 + 2 \frac{x}{b} + 3 \frac{x^2}{b^2} \right)
$$
$$
= \left(1 - 1\right) - 2 \frac{x}{b} - 2 \frac{x}{b} + \left(3 \frac{x^2}{b^2} - 3 \frac{x^2}{b^2}\right) = - 4 \frac{x}{b}
$$
Thus,
$$
\left[\left(1+ \frac{x}{b}\right)^{-2} - \left(1- \frac{x}{b}\right)^{-2}\right] \approx - 4 \frac{x}{b}
$$
Plug into (1);
$$
F_{\text{net}} = \frac{e^2}{4 \pi \epsilon_0 b^2}\left[- 4 \frac{x}{b}\right]
$$
$$
\boxed{F_{\text{net}} = \frac{-e^2\;x}{ \pi \epsilon_0 b^3}\;\hat i }
$$
$$\color{blue}{\bf [b]}$$
Recalling that the force of Hooke’s Law is given by $F=-kx$ where $k$ is the spring constant.
Applying that on the boxed formula above, then we got
$$
k = \frac{e^2}{ \pi \epsilon_0b^3}\tag 5
$$
Recalling that the angular frequency $ \omega $ of vibration is given by
$$
\omega = \sqrt{\frac{k}{m}} $$
Plugging from (5);
$$\omega= \sqrt{\frac{e^2}{ \pi \epsilon_0b^3 m}}
$$
Substitute the known;
$$
\omega = \sqrt{\frac{ (1.6 \times 10^{-19} )^2}{\pi (8.85\times 10^{-12})(0.30 \times 10^{-9} )^3 (27 \times 1.66 \times 10^{-27} )}}
$$
$$
\omega = 2.76 \times 10^{13} \; \text{rad/s}\tag 6
$$
We know that the ground state energy for a quantum harmonic oscillator is given by
$$
E_1 = \frac{1}{2} \hbar \omega
$$
So,
$$
E_1 = \frac{1}{2} (1.05 \times 10^{-34} ) (2.76 \times 10^{13} )=\bf 1.45 \times 10^{-21}\;\rm J
$$
$$
E_1 = \color{red}{\bf 0.00904 } \; \text{eV}
$$
By the same approach, higher energy levels are:
$$
E_2 = \frac{3}{2} (1.05 \times 10^{-34} ) (2.76 \times 10^{13} )=\bf 4.35 \times 10^{-21}\;\rm J
$$
$$
E_2 = \color{red}{\bf 0.0271 } \; \text{eV}
$$
$$
E_3 = \frac{5}{2} (1.05 \times 10^{-34} ) (2.76 \times 10^{13} )=\bf 7.24\times 10^{-21}\;\rm J
$$
$$
E_3 = \color{red}{\bf 0.0452 } \; \text{eV}
$$
$$
E_4 = \frac{7}{2} (1.05 \times 10^{-34} ) (2.76 \times 10^{13} )=\bf 1.01\times 10^{-20}\;\rm J
$$
$$
E_4 = \color{red}{\bf 0.0633 } \; \text{eV}
$$
$$\color{blue}{\bf [c]}$$
We can see that the difference between any two adjacent energy levels is given by
$$
\Delta E = E_2 - E_1 = E_3 - E_2 = E_4 - E_3 =\bf 0.0181\; \text{eV}
$$
So the wavelength $ \lambda $ of a photon with this energy is given by
$$
\lambda = \frac{hc}{\Delta E}
$$
Where $ h = 6.63 \times 10^{-34} \; \text{J} \cdot \text{s} $ and $ c = 3.0 \times 10^8 \; \text{m/s} $.
Substitute the known:
$$
\lambda = \frac{(6.63 \times 10^{-34}) (3.0 \times 10^8)}{(0.0181 \times 1.6 \times 10^{-19})} =\color{red}{\bf 69} \; \mu \text{m}
$$
which is in the infrared region.
