If the diameter of earth becomes half keeping mass to be constant, then the acceleration due to gravity at surface of earth becomes

The correct option is Four times.

Let's understand how the acceleration due to gravity on the surface of a spherical body like Earth is determined. The formula for the acceleration due to gravity ($g$) on the surface of the Earth of mass $M$ and radius $R$ is given by Newton's law of universal gravitation:
$g=\frac{GM}{R^2}$
where $G$ is the universal gravitational constant.

The diameter of the Earth is related to its radius by:
$D=2R$ or $R=\frac{D}{2}$
Thus, if the diameter becomes half, the radius of the Earth also becomes half of its original value. Let the new radius be $R^{\prime }$:
$R^{\prime }=\frac{R}{2}$

Since the mass of the Earth ($M$) remains constant, we can express the new acceleration due to gravity ($g^{\prime }$) as:
$g^{��}=\frac{GM}{{\left(R^{\prime }\right)}^2}$

Substituting $R^{\prime }=\frac{R}{2}$ into the equation:
$g^{\prime }=\frac{GM}{{\left(\frac{R}{2}\right)}^2}=\frac{GM}{\left(\frac{R^2}{4}\right)}=4\left(\frac{GM}{R^2}\right)$
Since $g=\frac{GM}{R^2}$ is the initial acceleration due to gravity, we get:
$g^{\prime }=4g$

Therefore, the acceleration due to gravity at the surface of the Earth becomes four times its original value.