1. Use conservation of angular momentum

If the Earth shrinks suddenly and no external torque acts, its angular momentum remains conserved:

$I_1 \omega_1 = I_2 \omega_2$

For a sphere, moment of inertia is proportional to

$I = kMR^2$

where $k$ is a constant, $M$ is mass, and $R$ is radius. Since mass remains the same,

$I \propto R^2$

2. Relate the new radius to the old radius

Given the new volume is $\dfrac{1}{64}$ of the original volume:

$\frac{V_2}{V_1} = \frac{1}{64}$

Since volume of a sphere is proportional to $R^3$,

$\left(\frac{R_2}{R_1}\right)^3 = \frac{1}{64}$

$\frac{R_2}{R_1} = \frac{1}{4}$

So,

$R_2 = \frac{R_1}{4}$

3. Find the change in moment of inertia

Because $I \propto R^2$,

$\frac{I_2}{I_1} = \left(\frac{R_2}{R_1}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$

Thus,

$I_2 = \frac{I_1}{16}$

4. Apply angular momentum conservation

$I_1 \omega_1 = I_2 \omega_2$

$I_1 \omega_1 = \frac{I_1}{16}\omega_2$

$\omega_2 = 16\omega_1$

5. Relate angular speed to time period

Since

$\omega = \frac{2\pi}{T}$

we have

$\omega \propto \frac{1}{T}$

Therefore,

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

Given $T_1 = 24$ h,

$T_2 = \frac{24}{16} \text{ h}$

Comparing with $\dfrac{24}{x}$ h,

$x = 16$

6. Comparison with stored answer

Derived answer: $16$

Stored correct answer: $16$

They match.