At the equator, a person stands on a spring balance. By what...

True weight of the person, $W = mg$

Apparent weight of the person at the equator due to Earth's rotation, $W' = m(g - \omega^2 R)$

Decrease in weight, $\Delta W = W - W' = m\omega^2 R$

Fractional decrease in weight is given by:

$\dfrac{\Delta W}{W} = \dfrac{m\omega^2 R}{mg} = \dfrac{\omega^2 R}{g}$

We know that the angular velocity of Earth is $\omega = \dfrac{2\pi}{T}$, where $T = 24 \times 3600 \text{ s}$.

Radius of Earth, $R \approx 6.4 \times 10^6 \text{ m}$ and $g = 9.8 \text{ m/s}^2$.

Substituting the values:

$\dfrac{\Delta W}{W} = \dfrac{4\pi^2 R}{T^2 g} = \dfrac{4 \times \pi^2 \times 6.4 \times 10^6}{(24 \times 3600)^2 \times 9.8}$

Using $\pi^2 \approx 9.87$, we get:

$\dfrac{\Delta W}{W} \approx \dfrac{252.67 \times 10^6}{73.15 \times 10^9} \approx 3.45 \times 10^{-3}$

This value is closest to $3.5 \times 10^{-3}$.