Correct Answer: $\frac{7}{8}$

Solution : Given: $2x+\frac{2}{x}=3$
$2(x+\frac{1}{x})=3$
$(x+\frac{1}{x})=\frac{3}{2}$
Cubing both sides, we get:
$(x+\frac{1}{x})^3=(\frac{3}{2})^3$
⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=\frac{27}{8}$
⇒ $x^3+\frac{1}{x^3}+3×\frac{3}{2}=\frac{27}{8}$
⇒ $x^3+\frac{1}{x^3}=\frac{27}{8}–\frac{9}{2}$
⇒ $x^3+\frac{1}{x^3}=–\frac{9}{8}$
To get the value of $x^{3}+\frac{1}{x^{3}}+2$, we need to add 2 both sides.
Thus, $x^{3}+\frac{1}{x^{3}}+2 = –\frac{9}{8}+2$
$\therefore x^{3}+\frac{1}{x^{3}}+2 = \frac{7}{8}$
Hence, the correct answer is $\frac{7}{8}$.