A layer of the ionosphere has 2.5 × 10¹¹ free electrons in each cubic metre…

Step 1. The critical frequency. $f_c = 9\sqrt{N} = 9 \times \sqrt{2.5 \times 10^{11}}\ \text{Hz} = 4.50\ \text{MHz}$.

Step 2. At $65^\circ$. $f_{\max} = \dfrac{f_c}{\cos i} = \dfrac{4.50\ \text{MHz}}{\cos 65^\circ} = 10.6\ \text{MHz}$.

Step 3. The smallest angle for $9.0$ MHz. We want $i$, so $\cos i = \dfrac{f_c}{f} = \dfrac{4.50\ \text{MHz}}{9.0\ \text{MHz}} = 0.500$, and $i = 60.0^\circ$.