Q.The value of acceleration due to gravity is maximum at (A) the equator of the Earth. (B) the centre of the Earth. (C) the pole of the Earth. (D) slightly above the surface of the Earth.

[!TLDR] Combining the Earth's rotation and its oblate (equator-bulging) shape, effective gravity is smallest at the equator and largest at the poles. [!ANSWER] (C) the pole of the Earth.

Two independent effects reduce g away from the poles. First, the Earth's rotation requires part of the gravitational pull to supply the centripetal force needed to keep a surface point moving in its daily circle; this reduction is $R\omega^2\cos^2\theta$, maximum at the equator ($\theta=0$) and exactly zero at the poles ($\theta=90^\circ$, since points on the rotation axis do not move in a circle at all). Second, the Earth is an oblate spheroid, bulging at the equator, so the equatorial radius exceeds the polar radius; since $g=GM/r^2$ falls with the square of distance from the centre, this larger radius further weakens g at the equator. Both effects point the same way, so measured g rises steadily from equator (about 9.7804 m/s^2) to pole (about 9.8322 m/s^2), the true maximum anywhere on the Earth's surface -- ruling out the equator, the centre (where g is exactly zero by the shell theorem), and a point above the surface (where g is always less than the surface value). [!ANSWER] (C) the pole of the Earth.

Compare the two effects (Earth's rotation and its oblate shape) that reduce effective gravity, both of which vanish at the poles, making the poles the point of maximum g.

Assuming g is simply constant everywhere on the surface, or confusing 'maximum g' with 'maximum weight felt', without accounting separately for rotation and shape.