Answer
See answer below.
Work Step by Step
The equilibrium for the full ionizations are:
$H_3BO_3 (aq)+3\ H_2O(l)\leftrightarrow 3\ H_3O^+(aq)\ +BO_3^{3-}(aq)$
$H_3PO_4 (aq)+3\ H_2O(l)\leftrightarrow 3\ H_3O^+(aq)\ +PO_4^{3-}(aq)$
Since both acids are in the same molar concentration and release the same amount of ions per molecule of acid, the difference in electrical conductivity is explained by the phosphoric acid reaction being more product-favored.
