A uniform solid cylinder of mass M and radius R has a wedge-shaped...

The angle of the removed sector is $60^{\circ}$.

The angle of the remaining portion of the cylinder is $360^{\circ} - 60^{\circ} = 300^{\circ}$.

Since the cylinder is uniform, the mass is proportional to the sector angle. The mass of the remaining portion $M'$ is:
$M' = \dfrac{300^{\circ}}{360^{\circ}} M = \dfrac{5}{6} M$

The moment of inertia of a solid cylinder (or any sector of it) about its central geometric axis is given by $\dfrac{1}{2} \times \text{mass} \times \text{radius}^2$.

Therefore, the moment of inertia of the remaining portion is:
$I' = \dfrac{1}{2} M' R^2 = \dfrac{1}{2} \left(\dfrac{5}{6} M\right) R^2 = \dfrac{5}{12} M R^2$

Answer: $\dfrac{5}{12}MR^2$