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**Hint:**The net acceleration imparted on particles by the combined influence of gravitation (from mass distribution within Earth) and centrifugal force (from the Earth's rotation) is denoted by the letter g. Such factors, such as the rotation of the Earth, add to the overall gravitational acceleration and, as a result, influence the weight of the material.

**Complete step by step answer:**

The acceleration of an object in free fall inside a vacuum is known as gravitational acceleration in physics (and thus without experiencing drag). This is the gradual increase in speed caused solely by gravitational attraction. All bodies accelerate in vacuum at the same rate at given GPS coordinates on the Earth's surface and a given altitude, independent of their masses or compositions; gravimetry is the calculation and study of these values.

We know that,

\[{\rm{g}} = {\rm{G}}\dfrac{M}{{{R^2}}}\]

$G$ = Gravitational Constant

\[G= 6.62 \times {10^{ - 11}}{\rm{N}}{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ - 2}}\]

$R$ = Radius of Earth Now

$M$ = Mass of Earth,

Now, the spaceship is at a distance $R’ = 2R$

So, $g$ will be:

\[{g^\prime } = G\dfrac{M}{{{R^{\prime 2}}}}\]

Substituting $R’ = 2R$

\[{{\rm{g}}^\prime } = {\rm{G}}\dfrac{M}{{4{R^2}}}\]

\[\Rightarrow {{\rm{g}}^\prime } = {\rm{G}}\dfrac{M}{{4{R^2}}} \\

\Rightarrow {{\rm{g}}^\prime } = \dfrac{g}{4} \\

\Rightarrow {{\rm{g}}^\prime } = \dfrac{{9.8}}{4} \\

\therefore {{\rm{g}}^\prime } = 2.45\;{\rm{m}}{{\rm{s}}^2}\]

**Hence option D is correct.**

**Note:**A non-rotating perfect sphere with equal mass density or whose density differs only with distance from the centre (spherical symmetry) will create a constant gravitational field at all points on its surface. The Earth rotates and is not spherically symmetric; instead, it is slightly flatter at the poles and bulges at the Equator, resulting in an oblate spheroid. As a result, there are minor variations in gravity magnitude across the surface.

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