Is this a violation of Equipartition theorem?
For a non-linear polyatomic molecule like methane, there are a total of $3N$ degrees of freedom, where $N = 4$ is the number of atoms in the molecule. This is broken down as follows:

$$\begin{array}{c|c} \hline
\text{Component} &amp; \text{# Degrees of freedom} \\
\hline
\text{Translational} &amp; 3 \\
\text{Rotational} &amp; 3 \\
\text{Vibrational} &amp; 3N - 6 \\ \hline
\end{array}$$

(Note that for a linear molecule, or for an atom, this breakdown is different.)

Under "normal" temperatures (close to room temperature) the vibrational contribution to the internal energy can be neglected. (The vibrational gap $\Delta E = \hbar\omega \gg k_\mathrm{B}T$ for "normal" $T$, so higher vibrational states are not populated and the equipartition theorem does not hold.) This is probably the reason why your book states that there are six total degrees of freedom.

The key is to realise that the kinetic energy of a molecule arises only from the translational degrees of freedom. From my limited knowledge of physics, it seems to be a common thing to separate translational degrees from the "internal" degrees of freedom, i.e. rotational, vibrational, and electronic. This allows the translational motion of a body to be modelled as the movement of a point mass, located at the centre of mass of the body.

In this paradigm, rotations and vibrations preserve the centre of mass of a molecule, and therefore don't contribute to the kinetic energy of the molecule. Therefore, the correct answer is $(1.5~\mathrm{mol})\cdot RT$.

Another way of looking at it is to consider the definition of the kinetic energy

$$E_k = \frac{1}{2}m\langle v^2\rangle = \frac{1}{2}m(\langle v_x^2\rangle + \langle v_y^2\rangle + \langle v_z^2\rangle)$$

where $v$ is the velocity of the centre of mass, $v_\mathrm{cm}$. For each quadratic term the contribution to the energy is $kT/2$, for a total of $3kT/2$ per molecule. These three terms correspond to precisely the three translational degrees of freedom.