Step 1: Understand orthogonal curves.
Two curves cut at right angles when the product of their slopes at the meeting point is $-1$. We need $a^2$.
Step 2: Slope of the ellipse.
Differentiate $\frac{x^2}{a^2}+\frac{y^2}{4}=1$: $\frac{2x}{a^2}+\frac{2y}{4}\frac{dy}{dx}=0$, so $m_1=-\frac{4x}{a^2 y}$.
Step 3: Slope of the second curve.
Differentiate $y^3=16x$: $3y^2\frac{dy}{dx}=16$, so $m_2=\frac{16}{3y^2}$.
Step 4: Set the orthogonal condition.
$m_1\cdot m_2=-1$ gives $\left(-\frac{4x}{a^2 y}\right)\left(\frac{16}{3y^2}\right)=-1$, i.e. $\frac{64x}{3a^2 y^3}=1$.
Step 5: Use the curve to remove y.
Since $y^3=16x$, replace $y^3$: $\frac{64x}{3a^2(16x)}=1$, which simplifies to $\frac{4}{3a^2}=1$.
Step 6: Solve for a squared.
Then $3a^2=4$, so $a^2=\frac{4}{3}$, which is option (D).
\[ \boxed{\,a^2=\tfrac{4}{3}\,} \]