Step 1: Use the orthogonal condition.
Two curves cut at right angles when the product of their slopes at the meeting point is $-1$.
Step 2: Slope of the ellipse.
From $\tfrac{x^2}{a^2}+\tfrac{y^2}{4}=1$, differentiating gives $\tfrac{2x}{a^2} + \tfrac{y}{2}\tfrac{dy}{dx}=0$, so $m_1 = -\tfrac{4x}{a^2 y}$.
Step 3: Slope of the cubic.
From $y^3 = 16x$, $3y^2\tfrac{dy}{dx} = 16$, so $m_2 = \tfrac{16}{3y^2}$.
Step 4: Set the product to $-1$.
\[ \left(-\frac{4x}{a^2 y}\right)\left(\frac{16}{3y^2}\right) = -1 \Rightarrow \frac{64x}{3a^2 y^3} = 1. \] Using $y^3 = 16x$, this is $\tfrac{64x}{3a^2(16x)} = \tfrac{4}{3a^2} = 1$, so $a^2 = \tfrac{4}{3}$.
\[ \boxed{a^2 = \tfrac{4}{3}} \]