Three solid spheres each of mass $$m$$ and diameter $$d$$ are stuck together such that the lines connecting the centres form an equilateral triangle of side of length $$d$$. The ratio $$\frac{I_0}{I_A}$$ of moment of inertia $$I_0$$ of the system about an axis passing the centroid and about center of any of the spheres $$I_A$$ and perpendicular to the plane of the triangle is:

For a solid sphere of mass $$m$$ and diameter $$d$$ (radius $$R = \frac{d}{2}$$), the moment of inertia about its center of mass axis is:  $$I_{\text{cm}} = \frac{2}{5}mR^2 = \frac{2}{5}m\left(\frac{d}{2}\right)^2 = \frac{m d^2}{10}$$

For the distance from the centroid $$O$$ of the equilateral triangle of side $$d$$ to the center of any sphere:

$$x = \frac{d}{\sqrt{3}}$$

For the total moment of inertia $$I_0$$ about an axis passing through the centroid $$O$$:

$$I_0 = 3 \times \left( I_{\text{cm}} + m x^2 \right) = 3 \times \left( \frac{m d^2}{10} + m \left(\frac{d}{\sqrt{3}}\right)^2 \right)$$

$$I_0 = 3 \times \left( \frac{m d^2}{10} + \frac{m d^2}{3} \right) = 3 \times \left(\frac{13}{30}\right) m d^2 = \frac{13}{10} m d^2$$

For the total moment of inertia $$I_A$$ about an axis passing through the center of sphere A:

$$I_A = I_{\text{cm}} + \left(I_{\text{cm}} + m d^2\right) + \left(I_{\text{cm}} + m d^2\right) = 3 I_{\text{cm}} + 2 m d^2$$

$$I_A = 3\left(\frac{m d^2}{10}\right) + 2 m d^2 = \frac{23}{10} m d^2$$

$$\frac{I_0}{I_A} = \frac{\frac{13}{10} m d^2}{\frac{23}{10} m d^2} = \frac{13}{23}$$