Correct Answer: 123

Solution : Given: $x+\frac{1}{x}=3$
$x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2=3^2-2=7$.
Similarly, $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3×x×\frac{1}{x}(x+\frac{1}{x})=3^3-3\times 3=18$.
Now, $x^5+\frac{1}{x^5}=(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})-(x+\frac{1}{x})$
Putting the values of $x^2+\frac{1}{x^2}$ and $x^3+\frac{1}{x^3}$, we get:
$\therefore x^5+\frac{1}{x^5}=7\times 18-3=123$
Hence, the correct answer is 123.