1. The problem states that $\sqrt[3]{\sqrt{x}} = 2$. We need to find the value of $x$ that satisfies this equation. 2. Rewrite the equation using exponents: $\sqrt{x} = x^{\frac{1}{2}}$ and $\sqrt[3]{y} = y^{\frac{1}{3}}$. So the equation becomes: $$\left(x^{\frac{1}{2}}\right)^{\frac{1}{3}} = 2$$ 3. Simplify the left side by multiplying the exponents: $$x^{\frac{1}{2} \times \frac{1}{3}} = x^{\frac{1}{6}} = 2$$ 4. To solve for $x$, raise both sides to the power of 6: $$\left(x^{\frac{1}{6}}\right)^6 = 2^6$$ $$x = 64$$ 5. Therefore, the value of $x$ is 64. Final answer: C 64