Answer: Option D
$$\eqalign{
& {\text{ }}x = 3 + 2\sqrt 2 \cr
& \Rightarrow {x^2} = {\left( {3 + 2\sqrt 2 } \right)^2} \cr
& \left( {{\text{Squaring both sides}}} \right) \cr
& \Rightarrow {x^2} = 9 + 8 + 12\sqrt 2 \cr
& \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr
& \Rightarrow \frac{1}{{{x^2}}} = \frac{1}{{17 + 12\sqrt 2 }} \times \frac{{17 - 12\sqrt 2 }}{{17 - 12\sqrt 2 }} \cr
& \Rightarrow \frac{1}{{{x^2}}} = 17 - 12\sqrt 2 \cr
& \therefore {\text{ }}{x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr
& = 17 + 12\sqrt 2 + 17 - 12\sqrt 2 \cr
& = 34 \cr} $$