This question assesses the understanding of bonding in electron-deficient compounds of Boron and Aluminium (Group 13) and their Lewis acidity.

Step 1: Analyze statement A

Statement A: $Al{\left( {C{H_3}} \right)_3}$ has the three-centre two-electron bonds in its dimeric structure.

1. Trimethylaluminium, $Al(CH_3)_3$, is an electron-deficient molecule because the central Aluminium atom has only 6 electrons in its valence shell.
2. To achieve stability, it dimerizes to form $Al_2(CH_3)_6$.
3. In the dimeric structure, there are four terminal methyl groups (forming normal 2-centre 2-electron bonds) and two bridging methyl groups.
4. Each bridging methyl group is bonded to two Aluminium atoms. An sp³ orbital from the bridging carbon atom overlaps simultaneously with sp³ orbitals from the two Aluminium atoms.
5. This forms a three-centre bond (Al-C-Al). The bond is formed by sharing a single pair of electrons among these three atoms. This type of bond is known as a three-centre two-electron (3c-2e) bond.
6. Therefore, the statement is correct.

Step 2: Analyze statement B

Statement B: $B{H_3}$ has the three-center two-electron bonds in its dimeric structure.

1. Borane, $BH_3$, is also an electron-deficient molecule as Boron has only 6 valence electrons.
2. It exists as a dimer, diborane ($B_2H_6$).
3. The structure of diborane has four terminal hydrogen atoms (in one plane) and two bridging hydrogen atoms (one above and one below the plane).
4. The terminal B-H bonds are normal 2-centre 2-electron (2c-2e) covalent bonds.
5. The bridging B-H-B bonds are three-centre two-electron (3c-2e) bonds. In these bonds, a single pair of electrons binds the three atoms (two Boron and one Hydrogen).
6. Therefore, the statement is correct.

Step 3: Analyze statement C

Statement C: $AlC{l_3}$ has the three-center two-electron bonds in its dimeric structure.

1. Aluminium trichloride, $AlCl_3$, is electron-deficient and dimerizes in the vapor phase to form $Al_2Cl_6$.
2. The structure of the dimer is similar to that of $Al_2(CH_3)_6$ with two bridging chlorine atoms and four terminal chlorine atoms.
3. However, the bonding in the bridge is different. Each bridging chlorine atom has lone pairs of electrons. It forms a normal covalent bond with one Aluminium atom and uses one of its lone pairs to form a coordinate (dative) bond with the other electron-deficient Aluminium atom.
4. Let's consider the $Al-Cl-Al$ bridge. It involves a 2-centre 2-electron covalent bond and a 2-centre 2-electron coordinate bond. In total, there are four electrons involved in the three-centre bridge. This is described as a three-centre four-electron (3c-4e) bond system.
5. Since the bond is a 3c-4e type, not 3c-2e, the statement is incorrect.

Step 4: Analyze statement D

Statement D: The Lewis acidity of $BC{l_3}$ is greater than that of $AlC{l_3}$.

1. Lewis acids are electron-pair acceptors. Both $BCl_3$ and $AlCl_3$ are Lewis acids because the central atoms (B and Al) have an incomplete octet and an empty p-orbital.
2. Several factors influence Lewis acidity. In $BCl_3$, there is $p\pi - p\pi$ back-bonding between the filled p-orbitals of chlorine and the empty 2p-orbital of boron. This delocalization of electrons reduces the electron deficiency on the boron atom, thereby decreasing its Lewis acidity.
3. In $AlCl_3$, back-bonding between the 3p-orbital of Al and the 3p-orbital of Cl ($3p\pi - 3p\pi$) is very weak and ineffective due to the larger size of the orbitals and greater energy difference.
4. Based on back-bonding alone, $AlCl_3$ should be a stronger Lewis acid. However, other factors like electronegativity and the energy released upon adduct formation must be considered.
5. Boron is significantly more electronegative than Aluminium (Pauling scale: B ≈ 2.04, Al ≈ 1.61). This makes the Boron atom in $BCl_3$ more electron-deficient and a better electron-pair acceptor than Aluminium in $AlCl_3$.
6. Experimental data, such as the enthalpy of formation of adducts with Lewis bases (e.g., pyridine), shows that $BCl_3$ forms more stable adducts than $AlCl_3$. This confirms that $BCl_3$ is a stronger Lewis acid than $AlCl_3$. The effect of higher electronegativity and smaller size of Boron outweighs the effect of back-bonding in this case.
7. Therefore, the statement is correct.

Conclusion

Based on the analysis, statements A, B, and D are correct, while statement C is incorrect.