1. Let the sum on two fair dice be $N$.

   Since two fair dice are rolled, the possible values of $N$ are: $2,3,4,5,6,7,8,9,10,11,12$

2. We need the probability that $N-2$, $\sqrt{3N}$, and $N+2$ are in geometric progression (G.P.).

   For three positive numbers $a,b,c$ to be in G.P., we must have: $b^2 = ac$

   Here, $a=N-2,\quad b=\sqrt{3N},\quad c=N+2$

   So, $(\sqrt{3N})^2 = (N-2)(N+2)$ $3N = N^2 - 4$ $N^2 - 3N - 4 = 0$ $N^2 - 4N + N - 4 = 0$ $(N-4)(N+1)=0$

   Hence, $N=4 \quad \text{or} \quad N=-1$

   Since $N$ is a sum of two dice, only $N=4$ is possible.

3. Now count the outcomes for which the sum is $4$.

   These are: $(1,3),(2,2),(3,1)$

   So number of favorable outcomes $=3$.

4. Total outcomes when two dice are rolled: $6\times 6 = 36$

   Therefore, the required probability is: $\frac{3}{36} = \frac{1}{12}$

5. It is given that this probability equals $\frac{k}{48}$.

   So, $\frac{k}{48} = \frac{1}{12}$ $k = 48\cdot \frac{1}{12} = 4$

6. Therefore, $k=4$

So the correct option is D.