1. Weight without rotation

If Earth is not rotating, the person's weight at the equator is $W = mg$ where $m$ is the mass of the person.

2. Weight when Earth rotates

At the equator, due to rotation, the required centripetal acceleration is $a_c = \omega^2 R$ So the apparent weight becomes $W' = m(g - \omega^2 R)$

Given that the apparent weight is $\dfrac{3}{4}W$, $m(g - \omega^2 R) = \frac{3}{4}mg$

3. Solve for $\omega$

Cancel $m$: $g - \omega^2 R = \frac{3}{4}g$ $\omega^2 R = g - \frac{3}{4}g = \frac{g}{4}$ $\omega^2 = \frac{g}{4R}$

Now substitute: $g = 10\ \text{m/s}^2, \qquad R = 6400\ \text{km} = 6.4 \times 10^6\ \text{m}$

So, $\omega^2 = \frac{10}{4 \times 6.4 \times 10^6}$ $\omega^2 = \frac{10}{25.6 \times 10^6}$ $\omega^2 = 3.90625 \times 10^{-7}$

Therefore, $\omega = \sqrt{3.90625 \times 10^{-7}}$ $\omega = 6.25 \times 10^{-4}\ \text{rad/s}$

4. Match with options

$6.25 \times 10^{-4} = 0.625 \times 10^{-3}\ \text{rad/s}$

This matches:

Option C: $0.63 \times 10^{-3}\ \text{rad/s}$

5. Final answer

The Earth must rotate with angular speed $\boxed{0.63 \times 10^{-3}\ \text{rad/s}}$

So the correct option is C.