I presume that you have learned that the force between charge Q and charge Q', with distance r between them is $\displaystyle k_e\frac{QQ'}{r^2}$ where $\displaystyle k_e$ is "Coulomb's constant" (about $\displaystyle 9\times 10^9$ Newton meters squared per Coulomb squared).
Imagine dividing the ring into many sectors by drawing lines, a constant angle $\displaystyle \Delta\theta$ apart so that the length of each sector is $\displaystyle R\Delta\theta$ where R is the radius of the ring. Similarly imagine dividing the rod into many short segments of length $\displaystyle \Delta x$. Given point P on the ring, at angle $\displaystyle \theta$ and point Q on the rod at height x, The distance between P and Q is, by the Pythagorean theorem, $\displaystyle r= \sqrt{x^2+ R^2}$. That little segment of the ring is the fraction $\displaystyle \frac{R\Delta\theta}{2\pi R}= \frac{\Delta\theta}{2\pi}$ of the ring so (assuming the charge is uniformly distributed) the charge on it is $\displaystyle \frac{Q\Delta\theta}{2\pi}$. That little segment of the bar is the fraction $\displaystyle \frac{\Delta x}{3\alpha 2\alpha}= \frac{\Delta x}{\alpha}$ so (assuming the charge is uniformly distributed, the charge on it is $\displaystyle \frac{Q\Delta x}{\alpha}$.
Putting those together, the force between P and Q is $\displaystyle k_e\frac{Q^2\Delta x\Delta\theta}{2\pi\alpha(x^2+ R^2)}$.
The total force between ring and rod is the sum of those over all segments of the ring and rod. Taking the limit as $\displaystyle \Delta\theta$ and $\displaystyle \Delta x$ become infinitesimal, we get the double integral
$\displaystyle \frac{k_eQ^2}{2\pi\alpha}\int_{x= \alpha}^{2\alpha}\int_{\theta= 0}^{2\pi}\frac{dxd\theta}{x^2+ R^2}$.
