Given data

• Mass of each sphere: $m = 2\,\text{kg}$
• Radius of each sphere: $R = 50\,\text{cm} = 0.5\,\text{m}$
• Distance between centers: $d = 150\,\text{cm} = 1.5\,\text{m}$

Axis of rotation The axis is perpendicular to the rod and passes through its midpoint.

Hence, the distance of each sphere’s center from the axis is $a = \frac{1.5}{2} = 0.75\,\text{m}.$

Moment of inertia of one sphere about the given axis For a solid sphere, moment of inertia about any diameter through its center is $I_{\text{cm}} = \frac{2}{5}mR^2.$

Substituting values: $I_{\text{cm}} = \frac{2}{5}(2)(0.5)^2 = \frac{4}{5}\cdot 0.25 = 0.2\,\text{kg m}^2.$

Using the parallel axis theorem, $I = I_{\text{cm}} + ma^2$ $= 0.2 + 2(0.75)^2$ $= 0.2 + 2(0.5625)$ $= 0.2 + 1.125 = 1.325\,\text{kg m}^2.$

Moment of inertia of both spheres $I_{\text{total}} = 2 \times 1.325 = 2.65\,\text{kg m}^2.$

Compare with given form It is given that $I_{\text{total}} = \frac{x}{20}\,\text{kg m}^2.$

So, $\frac{x}{20} = 2.65$ $x = 2.65 \times 20 = 53.$

Final answer $\boxed{x = 53}$