1. Initial situation

   The two identical conducting spheres $P$ and $S$ each have charge $Q$.

   So the initial force between them is $F = k\frac{Q\cdot Q}{r^2} = k\frac{Q^2}{r^2} = 16\,\text{N}$

2. First contact: sphere $R$ touches $P$

   Sphere $R$ is identical and initially uncharged.

   Before contact:

   • Charge on $P = Q$
   • Charge on $R = 0$

   Since the spheres are identical, after contact the total charge $Q$ is shared equally: $P = \frac{Q}{2}, \qquad R = \frac{Q}{2}$

3. Second contact: sphere $R$ touches $S$

   Now before contact:

   • Charge on $R = \frac{Q}{2}$
   • Charge on $S = Q$

   Total charge during contact: $Q + \frac{Q}{2} = \frac{3Q}{2}$

   Since $R$ and $S$ are identical, after contact each gets half: $R = \frac{3Q}{4}, \qquad S = \frac{3Q}{4}$

   Sphere $P$ remains unchanged at: $P = \frac{Q}{2}$

4. New force between $P$ and $S$

   Final charges:

   • On $P$: $\frac{Q}{2}$
   • On $S$: $\frac{3Q}{4}$

   Therefore, $F' = k\frac{\left(\frac{Q}{2}\right)\left(\frac{3Q}{4}\right)}{r^2}$ $F' = \frac{3}{8}k\frac{Q^2}{r^2}$

   But $k\frac{Q^2}{r^2} = 16$

   So, $F' = \frac{3}{8}\times 16 = 6\,\text{N}$

5. Checking options

   • A: $1\,\text{N}$ ❌
   • B: $6\,\text{N}$ ✅
   • C: $12\,\text{N}$ ❌
   • D: $4\,\text{N}$ ❌

Therefore, the correct answer is Option B.