Answer: Option D
$$\eqalign{
& \frac{{{x^{24}} + 1}}{{{x^{12}}}} = 7{\text{ }}\left( {{\text{Given}}} \right) \cr
& \Rightarrow \frac{{{x^{24}}}}{{{x^{12}}}} + \frac{1}{{{x^{12}}}} = 7 \cr
& \Rightarrow {x^{12}} + \frac{1}{{{x^{12}}}} = 7 \cr
& {\text{Cubing both sides}} \cr
& \Rightarrow {\left( {{x^{12}} + \frac{1}{{{x^{12}}}}} \right)^3} = {7^3} \cr} $$
  $$ \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} + \frac{{3 \times {x^{12}} \times 1}}{{{x^{12}}}}$$     $$\left( {{x^{12}} + \frac{1}{{{x^{12}}}}} \right) = $$    $$343$$
$$\eqalign{
& \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} + 3\left( 7 \right) = 343 \cr
& \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} = 343 - 21 \cr
& \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} = 322 \cr
& \Rightarrow \frac{{{x^{72}} + 1}}{{{x^{36}}}} = 322 \cr} $$