1. Count total valence electrons in ${\rm I}_3^-$:

   • Each iodine atom has $7$ valence electrons.
   • For $3$ iodine atoms: $3 \times 7 = 21$
   • One extra electron due to the negative charge: $21 + 1 = 22$

   So, total valence electrons $= 22$.

2. Draw the basic structure:

   The ion is linear with one central iodine: $\mathrm{I-I-I}$

   There are two single bonds.

3. Count electrons used in bonding:

   • Each single bond contains $2$ electrons.
   • Two bonds use $2 \times 2 = 4$ electrons.

   Remaining electrons: $22 - 4 = 18$

4. Convert remaining electrons into lone pairs:

   • Each lone pair contains $2$ electrons.
   • Therefore, number of lone pairs: $\frac{18}{2} = 9$

5. Check distribution:

   • Each terminal iodine has $3$ lone pairs.
   • Central iodine also has $3$ lone pairs.

   Total lone pairs: $3 + 3 + 3 = 9$

6. Evaluate options:

   • A: $3$ ❌
   • B: $9$ ✅
   • C: $6$ ❌
   • D: $12$ ❌

Therefore, the total number of lone pairs in ${\rm I}_3^-$ is 9.