B

Step 1: Given data: Height of the diamond block, h = 12 mm; refractive index of diamond, n_{diamond} = 2.4; diameter of the circular spot, D = 18 mm.
Step 2: The radius of the circular spot is R = \frac{D}{2} = \frac{18 \text{ mm}}{2} = 9 \text{ mm}.
Step 3: The critical angle for diamond to liquid interface can be obtained using Snell's law. At critical angle, \( n_{diamond} \sin(\theta_c) = n_{liquid} \sin(90^\circ) \). This leads to \( \sin(\theta_c) = \frac{n_{liquid}}{n_{diamond}} \).
Step 4: The radius R can also be expressed in terms of the height h and the critical angle. Hence, using geometry: \( R = h \tan(\theta_c) \). Therefore, \( \tan(\theta_c) = \frac{R}{h} = \frac{9 \text{ mm}}{12 \text{ mm}} = \frac{3}{4}. \)
Step 5: Find the critical angle: \(\tan(\theta_c) = \frac{3}{4} \Rightarrow \theta_c = \tan^{-1}(\frac{3}{4}) \).
Step 6: Using the identity \( \sin^2(\theta_c) + \cos^2(\theta_c) = 1 \) to find \( n_{liquid} = n_{diamond} \sin(\theta_c) \). Calculate \( n_{liquid} \approx 1.6.
Therefore, Option B is the correct answer.