1. **State the problem:** We need to find the value of $\cos E$ in a right triangle $EFG$ where the right angle is at vertex $F$. The sides adjacent to $F$ are $EF = 18$ and $FG = 22$.\n\n2. **Recall the definition of cosine in a right triangle:** $\cos$ of an angle is the ratio of the length of the adjacent side to the hypotenuse. For angle $E$, the adjacent side is $EF$, and the hypotenuse is $EG$.\n\n3. **Find the hypotenuse $EG$ using the Pythagorean theorem:**\n$$EG = \sqrt{EF^2 + FG^2} = \sqrt{18^2 + 22^2} = \sqrt{324 + 484} = \sqrt{808}$$\n\n4. **Simplify $\sqrt{808}$ if possible:**\n$$808 = 4 \times 202 \Rightarrow \sqrt{808} = \sqrt{4 \times 202} = \sqrt{4} \times \sqrt{202} = 2\sqrt{202}$$\n\n5. **Calculate $\cos E$:**\n$$\cos E = \frac{EF}{EG} = \frac{18}{2\sqrt{202}}$$\n\n6. **Simplify the fraction:**\n$$\cos E = \frac{18}{2\sqrt{202}} = \frac{\cancel{18}}{\cancel{2}\sqrt{202}} = \frac{9}{\sqrt{202}}$$\n\n7. **Rationalize the denominator:**\n$$\cos E = \frac{9}{\sqrt{202}} \times \frac{\sqrt{202}}{\sqrt{202}} = \frac{9\sqrt{202}}{202}$$\n\n8. **Calculate the decimal value:**\n$$\sqrt{202} \approx 14.2127$$\n$$\cos E \approx \frac{9 \times 14.2127}{202} = \frac{127.9143}{202} \approx 0.633$$\n\n9. **Round to the nearest hundredth:**\n$$\cos E \approx 0.63$$\n\n**Final answer:** $\boxed{0.63}$