Interpret the geometry

We have:

• A rod of length $2R$ and mass $M$
• Two identical solid spherical balls, each of mass $M$ and radius $R$, attached at the two ends of the rod
• Axis: passing through the centre of the rod and perpendicular to the rod

Since each sphere of radius $R$ is attached at an end of the rod, the distance of each sphere's centre from the rod's centre is $d = R + \frac{2R}{2} = R + R = 2R.$

Moment of inertia of the rod about the given axis

For a uniform rod of length $2R$ about an axis through its centre and perpendicular to its length, $I_{\text{rod}} = \frac{1}{12} M(2R)^2 = \frac{1}{3}MR^2.$

Moment of inertia of one sphere about the given axis

For one solid sphere:

• Moment of inertia about its own centre: $I_{\text{cm}} = \frac{2}{5}MR^2.$
• Distance of its centre from the given axis: $2R$

By parallel axis theorem, $I_{\text{one sphere}} = I_{\text{cm}} + Md^2 = \frac{2}{5}MR^2 + M(2R)^2.$ $I_{\text{one sphere}} = \frac{2}{5}MR^2 + 4MR^2 = \frac{22}{5}MR^2.$

Moment of inertia of two spheres

$I_{\text{two spheres}} = 2\times \frac{22}{5}MR^2 = \frac{44}{5}MR^2.$

Total moment of inertia of the system

$I_{\text{total}} = I_{\text{rod}} + I_{\text{two spheres}}$ $I_{\text{total}} = \frac{1}{3}MR^2 + \frac{44}{5}MR^2$

Taking LCM $15$, $I_{\text{total}} = \left(\frac{5}{15} + \frac{132}{15}\right)MR^2 = \frac{137}{15}MR^2.$

Match with options

$\boxed{I = \frac{137}{15}MR^2}$

So the correct option is B.