Step-by-step Solution:

1. Understand the Setup and Governing Principle: The problem describes a point source of light at the bottom of a denser medium (a block) immersed in a rarer medium (a liquid). The light rays from the source travel from the denser medium to the rarer medium. A circular bright spot is formed on the top surface because light rays incident at an angle less than the critical angle ($C$) get refracted into the liquid, while rays incident at an angle greater than the critical angle undergo Total Internal Reflection (TIR) and are reflected back into the block. The edge of this bright spot is formed by the rays that are incident at exactly the critical angle, $C$. These rays get refracted at $90^\circ$ and travel along the surface.

2. Relate Geometry to the Critical Angle: Let's analyze the geometry. Consider a ray from the source S that strikes the edge of the bright spot, P, on the top surface. Let O' be the point on the top surface directly above S. Then, triangle SO'P is a right-angled triangle.

   • The height of the block, $h = SO' = 10$ mm.
   • The diameter of the bright spot is $d = 11.54$ mm.
   • The radius of the bright spot is $r = O'P = d/2 = 11.54 / 2 = 5.77$ mm.
   • The angle of incidence for the ray SP at point P is the angle $\angle O'SP$. This angle is equal to the critical angle, $C$. From the right-angled triangle SO'P, we can write the trigonometric relation: $\tan(C) = \frac{\text{opposite}}{\text{adjacent}} = \frac{O'P}{SO'} = \frac{r}{h}$

3. Calculate the Critical Angle: Substituting the given values: $\tan(C) = \frac{5.77 \text{ mm}}{10 \text{ mm}} = 0.577$ We can recognize that $0.577$ is a very good approximation for $1/\sqrt{3}$. $\tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.57735$ Therefore, we can conclude that the critical angle is $C = 30^\circ$.

4. Apply Snell's Law for Critical Angle: Let the refractive index of the block be $n_1 = 2.72$ and the refractive index of the liquid be $n_2$. According to Snell's law, at the critical angle of incidence $C$, the angle of refraction is $90^\circ$. $n_1 \sin(C) = n_2 \sin(90^\circ)$ Since $\sin(90^\circ) = 1$, the equation simplifies to: $n_2 = n_1 \sin(C)$

5. Calculate the Refractive Index of the Liquid ($n_2$): We have $n_1 = 2.72$ and we found $C = 30^\circ$. We know that $\sin(30^\circ) = 0.5$. Substituting these values into the equation: $n_2 = 2.72 \times \sin(30^\circ)$ $n_2 = 2.72 \times 0.5$ $n_2 = 1.36$

6. Conclusion: The calculated refractive index of the liquid is 1.36. This matches option C.

   Alternative Check (without assuming $C=30^\circ$): From $\tan(C) = 0.577$, we can find $\sin(C)$ using the identity $\sin(C) = \frac{\tan(C)}{\sqrt{1+\tan^2(C)}}$. $\sin(C) = \frac{0.577}{\sqrt{1 + (0.577)^2}} = \frac{0.577}{\sqrt{1 + 0.332929}} = \frac{0.577}{\sqrt{1.332929}} \approx \frac{0.577}{1.1545} \approx 0.4998 \approx 0.5$ This confirms that our assumption was correct. $n_2 = n_1 \sin(C) = 2.72 \times 0.5 = 1.36$