Correct Answer: $24\sqrt2$

Solution : Given: $x=\sqrt{2}+1$
Thus, $\frac{1}{x}=\sqrt{2}-1$
$x+\frac{1}{x}=(\sqrt{2}+1)+(\sqrt{2}-1)=(\sqrt{2}+1+\sqrt{2}-1)=2\sqrt{2}$
$x-\frac{1}{x}=(\sqrt{2}+1)-(\sqrt{2}-1)=(\sqrt{2}+1-\sqrt{2}+1)=2$
So, $(x^2+\frac{1}{x^2})=6$
We know that,
$x^4-\frac{1}{x^4}=(x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})=(x^2+\frac{1}{x^2})(x+\frac{1}{x})(x-\frac{1}{x})$
Putting the values we get,
⇒ $x^4-\frac{1}{x^4}=6×2×2\sqrt{2}$
$\therefore x^4-\frac{1}{x^4}=24\sqrt{2}$
Hence, the correct answer is $24\sqrt{2}$.