The apparent weight of a person on the equator (Lattitude $ \alpha =0$) is given by
$W'=W-mR_ew^2$
$g'=g -R_e w^2$
$g'=0$
$g-Rw^2=0$
$w=\sqrt {\large\frac{g}{R}}$
$\quad=\sqrt {\large\frac{10}{6400 \times 10^3}}$
$\quad=1.25 \times 10^{-3} rad /s$
Hence a is the correct answer.