The rotation through the center in the plane of the cylinder just means that the rotation is through the \(\displaystyle x\)-axis or the \(\displaystyle y\)-axis,
NOT through the \(\displaystyle z\)-axis.
There is a beautiful theorem which is called the Perpendicular-axis Theorem. It states that the sum of the moments of inertia of a body through the plane axes equal to the moment of inertia through its center.
In other words, this theorem says this:
\(\displaystyle I_z = I_x + I_y\)
We notice that the moment of inertia through the \(\displaystyle x\)-axis is symmetric to that through the \(\displaystyle y\)-axis.
Then,
\(\displaystyle I_z = I_x + I_y = I_x + I_x = I_y + I_y = 2I_x = 2I_y\)
Here is the crucial part. The moment of inertia through the center of a uniform cylinder, a thin uniform cylinder, a uniform disc, a uniform coin, or whatever similar uniform shape is the same as long as the mass is distributed evenly through the circle area.
Therefore, the moment of inertia of a thin uniform cylinder through its center is: \(\displaystyle I_z = \frac{1}{2}MR^2\).
If you are confused why this result is true by saying the uniform cylinder has a height \(\displaystyle h\) while the thin uniform cylinder doesn't have a height, or it might have a very tiny one (like the coin), you can always prove this kind of results by using the definition of moment of inertia, that is: \(\displaystyle I = \int R^2 \ dm\).
Finally, we can answer the op problem.
The moment of inertia through the center in the plane of the cylinder is:
\(\displaystyle I_x = I_y = \frac{1}{2}I_z = \frac{1}{2}\frac{1}{2}MR^2 = \textcolor{blue}{\frac{1}{4}MR^2}\)
