Correct Answer: $47$

Solution : Given: $x^2-3x+1=0$
Divide both sides by $x$, we get,
$⇒ x-3+\frac{1}{x}=0$
$\therefore x+\frac{1}{x}=3$
Squaring both sides, we get,
$⇒(x+\frac{1}{x})^2=3^2$
$⇒x^2+\frac{1}{x^2}+2×x×\frac{1}{x}=9$
$\therefore x^2+\frac{1}{x^2}=7$
Again squaring both sides, we get,
$⇒(x^2+\frac{1}{x^2})^2=7^2$
$⇒x^4+\frac{1}{x^4}+2×x^2×\frac{1}{x^2}=49$
$⇒(x^4+\frac{1}{x^4})=49-2$
$\therefore(x^4+\frac{1}{x^4})=47$
Hence, the correct answer is $47$.