A particle of mass $$m$$ is under the influence of the gravitational field of a body of mass $$M$$ ($$\gg m$$). The particle is moving in a circular orbit of radius $$r_0$$ with time period $$T_0$$ around the mass $$M$$. Then, the particle is subjected to an additional central force, corresponding to the potential energy $$V_c(r) = m\alpha / r^3$$, where $$\alpha$$ is a positive constant of suitable dimensions and $$r$$ is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius $$r_0$$ in the combined gravitational potential due to $$M$$ and $$V_c(r)$$, but with a new time period $$T_1$$, then $$(T_1^2 - T_0^2)/T_1^2$$ is given by

[$$G$$ is the gravitational constant.]

The gravitational potential energy due to the central mass $$M$$ is
$$V_g(r)= -\dfrac{G M m}{r}\,.$$

For a circular orbit of radius $$r_0$$ under gravity alone, the inward (centripetal) force balance is
$$\dfrac{m v_0^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}} \;\;\Longrightarrow\;\; v_0^{\,2}= \dfrac{G M}{r_0}\,.$$

The corresponding time period is therefore
$$T_0 =\dfrac{2\pi r_0}{v_0}=2\pi\sqrt{\dfrac{r_0^{3}}{G M}} \;\;\Longrightarrow\;\; T_0^{2}= \dfrac{4\pi^{2} r_0^{3}}{G M}\,.$$

Now an additional central potential is applied:
$$V_c(r)=\dfrac{m\alpha}{r^{3}}\qquad (\alpha\gt 0).$$

The radial force associated with this potential is obtained from $$F_r=-\dfrac{dV_c}{dr}$$:
$$F_c(r)=-\frac{d}{dr}\left(\dfrac{m\alpha}{r^{3}}\right)=+\,\dfrac{3m\alpha}{r^{4}}.$$
Since this force is positive in the outward radial direction, it is repulsive. Hence it opposes gravitation.

To keep the particle in the same circular orbit of radius $$r_0$$, the inward gravitational pull must now be reduced by this repulsion:
$$\dfrac{m v_1^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}}-\dfrac{3m\alpha}{r_0^{4}} \;\;\Longrightarrow\;\; v_1^{\,2}= \dfrac{G M}{r_0}-\dfrac{3\alpha}{r_0^{3}}.$$

The new time period becomes
$$T_1 =\dfrac{2\pi r_0}{v_1}\,,\qquad T_1^{2}= \dfrac{4\pi^{2} r_0^{2}}{v_1^{\,2}} =\dfrac{4\pi^{2} r_0^{3}} {\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$$ Define $$A=4\pi^{2}r_0^{3}$$ for brevity, so that
$$T_0^{2}= \dfrac{A}{G M},\qquad T_1^{2}= \dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$$

We now evaluate the required ratio:
$$\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}} =1-\dfrac{T_0^{2}}{T_1^{2}} =1-\dfrac{\dfrac{A}{G M}} {\dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}} =1-\dfrac{G M-\dfrac{3\alpha}{r_0^{2}}}{G M} =\dfrac{3\alpha}{G M r_0^{2}}\;.$$

Thus
$$\boxed{\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}}=\dfrac{3\alpha}{G M r_0^{2}}}\,.$$
Hence the correct option is:
Option A which is: $$\dfrac{3\alpha}{G M r_0^{2}}.$$