Step 1: Use the fundamental relation between a matrix and its adjoint: $A \cdot \operatorname{adj}(A) = \det(A)\, I_n$, where $I_n$ is the identity matrix.
Step 2: Take determinants of both sides. $$\det(A)\cdot\det(\operatorname{adj} A) = \det(\det(A)\, I_n)$$ For an $n\times n$ scalar matrix $cI_n$, $\det(cI_n) = c^n$. So the right side equals $[\det(A)]^n$.
Step 3: With $n=3$ and $B = \operatorname{adj}(A)$, this gives $$\det(A)\cdot\det(B) = [\det(A)]^3$$ Substitute $\det(B) = 64$: $$64\det(A) = [\det(A)]^3$$
Step 4: Rearranging, $[\det(A)]^3 - 64\det(A) = 0$, i.e. $\det(A)\left([\det(A)]^2 - 64\right) = 0$. So either $\det(A) = 0$ or $\det(A) = \pm 8$.
Step 5: If $\det(A) = 0$, then $A$ is singular, and for a $3\times3$ matrix this forces $\operatorname{adj}(A)$ to have rank at most $1$, making $\det(\operatorname{adj} A) = 0$. This contradicts the given $\det(B) = 64 \neq 0$. So $\det(A) = 0$ must be rejected.
\[\boxed{\det(A) = \pm 8}\]