**Solution:**

We first calculate the wavenumber $k$ at $5 \mathrm{GHz}$
$$
k=\frac{2 \pi f \sqrt{\epsilon_r}}{c}=157.08 / m
$$

The dominant mode is $\mathrm{TE}_{101}$ so $m=1, n=0$. Then we can find the resonance for $l=1$ and 2

$$
d=\frac{l \pi}{\sqrt{k^2-\left(\frac{\pi}{a}\right)^{\wedge} 2}}
$$

For $l=1$,

$$
d=2.20 \mathrm{~cm}
$$

For $l=2$,

$$
d=4.40 \mathrm{~cm}
$$

Now we can calculate the unloaded Q. The Q due to conductor loss is given by,

$$
Q_c=\frac{(k a d)^3 b \eta}{2 \pi^2 R_s} \frac{1}{\left(2 l^2 a^3 b+2 b d^3+l^2 a^3 d+a d^3\right)}
$$

Where $\eta=\frac{377}{\sqrt{\epsilon_r}}=251.3 \Omega$ for polyethylene.

For $l=1, \quad Q_c=8,403$

For $l=2, \quad Q_c=11,898$

The Q due to dielectric loss is given by,

$$
Q_d=\frac{1}{\tan \delta}=2,500
$$

For both $l=1$ and 2. Then the total unloaded $Q s$ are

For $l=1, \quad Q_0=\left(\frac{1}{8,403}+\frac{1}{2,500}\right)^{-1}=1927$

For $l=2, \quad Q_0=\left(\frac{1}{11,898}+\frac{1}{2,500}\right)^{-1}=2065$