= Solution
Let the uniform <mass density> be $\rho=3M/(4\pi a^3)$. In <spherical polar coordinates> about the rotation axis, the perpendicular distance is $r\sin\theta$ and the volume element is $r^2\sin\theta\,dr\,d\theta\,d\phi$. The <moment of inertia> is therefore
$$
I=\rho\int_0^a r^4\,dr\int_0^\pi\sin^3\theta\,d\theta\int_0^{2\pi}d\phi=\frac{8\pi\rho a^5}{15}=\boxed{\frac25Ma^2}.
$$
In the given <rigid body> <kinetic energy> decomposition, $\tfrac12M|\dot{\mathbf R}|^2$ is the translational <kinetic energy> of the whole mass moving at its <center of mass> velocity. The term $\tfrac12I\omega^2$ is the rotational <kinetic energy> about the <center of mass>. The mixed term vanishes because the mass-weighted relative positions sum to zero.
For <rolling without slipping>, the point of contact has zero instantaneous velocity relative to the stationary surface. Its rotational velocity is opposite the <center of mass> velocity and has magnitude $a|\omega|$. Hence $V=a|\omega|$. Substituting this and the <moment of inertia of a uniform solid sphere> gives
$$
\boxed{T=\frac12MV^2+\frac12\left(\frac25Ma^2\right)\frac{V^2}{a^2}=\frac7{10}MV^2.}
$$
The rotational axis is through the <center of mass> and parallel to the surface, perpendicular to the direction of rolling.
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