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The solubility of $PbC{l_2}$ at ${25\,^o}C$ is $6.3 \times {10^{ - 3}}$ $mole/litre$. Its solubility product at that temperature is

• A
  $(6.3 \times {10^{ - 3}}) \times (6.3 \times {10^{ - 3}})$
• B
  $(6.3 \times {10^{ - 3}}) \times (12.6 \times {10^{ - 3}})$
• ✓
  $(6.3 \times {10^{ - 3}}) \times {(12.6 \times {10^{ - 3}})^2}$
• D
  $(12.6 \times {10^{ - 3}}) \times (12.6 \times {10^{ - 3}})$