From the conservation of angular momentum, we have:

$ \frac{2}{5}MR^2\omega_1 = \frac{2}{5}M\left(\frac{R}{4}\right)^2\omega_2 $

This simplifies to:

$ MR^2\omega_1 = \frac{MR^2}{16}\omega_2 $

From this, we can derive the ratio of the initial and final angular velocities:

$ \frac{\omega_1}{\omega_2} = \frac{1}{16} $

Since the angular velocity (\omega) is inversely proportional to the period of rotation (T) ((\omega = \frac{2\pi}{T})), we can write:

$ \frac{T_2}{T_1} = \frac{1}{16} $

We can express this ratio in terms of the variable (x):

$ \frac{T_1}{T_2} = \frac{16}{1} = \frac{24}{x} $

Solving this equation for (x) gives:

$ x = 16 $

So, if the Earth suddenly shrinks to ( $\frac{1}{64}$ )th of its original volume with its mass remaining the same, the period of rotation of Earth becomes ( $\frac{24}{16}$ )h, or 1.5 hours. Therefore, the value of (x) is 16.