Where will it be profitable to purchase 1 kg sugar (by spring balance)

The correct option is At equator.

Let us understand the physics behind this step-by-step.
A spring balance measures the weight of an object, which is the gravitational force acting on it. The weight $W$ of an object is given by the formula:
$W=mg$
where $m$ is the mass of the object and $g$ is the acceleration due to gravity at that location.

Due to the rotation of the Earth and its oblate spheroidal shape (being flattened at the poles and bulging at the equator), the effective acceleration due to gravity $g$ varies with latitude $\theta$. The relationship is given by:
$g_{\theta }=g-{\omega }^{2}R{\cos }^2\theta$
where:
- $g$ is the acceleration due to gravity without rotation,
- $\omega$ is the angular velocity of the Earth's rotation,
- $R$ is the radius of the Earth, and
- $\theta$ is the latitude.

At the poles ($\theta =90°$), the term $\cos 90°=0$, so the acceleration due to gravity is maximum:
$g_{\mathrm{pole}}=g$

At the equator ($\theta =0°$), the term $\cos 0°=1$, so the acceleration due to gravity is minimum:
$g_{\mathrm{equator}}=g-{\omega }^{2}R$

Since a spring balance calibrated at a standard location measures force, it will show 1 kg of sugar when the force equals the weight of 1 kg mass under standard gravity. Since $g_{\mathrm{equator}}<g_{\mathrm{pole}}$, the gravitational pull per unit mass is weakest at the equator. Consequently, to get a reading of 1 kg on the spring balance at the equator, you actually need a larger actual mass (more quantity) of sugar compared to what you would need at the poles. Since you pay for the reading on the balance (1 kg) but receive more actual mass of sugar, it is most profitable to purchase sugar at the equator.