= Solution
\b[$I_{C_4}=4m_B R^2(1-\cos\theta)$; the central atom makes no contribution.] Let $\rho$ be the radius of the square of outer atoms and $h_A$ the central atom's height above its plane. Then $R^2=\rho^2+h_A^2$. Adjacent bond vectors have perpendicular in-plane components, so
$$
\cos\theta=\frac{h_A^2}{R^2},\qquad \rho^2=R^2(1-\cos\theta).
$$
The <centre of mass> lies on the fourfold axis, even when $m_A\ne m_B$. Thus the central atom and the shift of origin along the axis contribute no perpendicular lever arm. The four outer atoms each contribute $m_B\rho^2$, giving
$$
\boxed{I_{C_4}=m_A(0)^2+4m_B\rho^2=4m_BR^2(1-\cos\theta).}
$$
The result concerns the smallest adjacent bond angle; the angle between opposite outer atoms is a different angle in a pyramid.
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