The given question involves understanding the boiling points of Group 16 hydrides and the effect of molecular mass and hydrogen bonding on these boiling points.

Statement I claims that the boiling points of Group 16 hydrides follow this order:

$$\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{S}.$$

This statement is indeed true. Water ($\mathrm{H}_2 \mathrm{O}$) has a much higher boiling point compared to the other hydrides in Group 16. This anomaly primarily arises due to the extensive hydrogen bonding present in water, which greatly increases its boiling point. In the absence of such strong hydrogen bonding, heavier molecules like $\mathrm{H}_2 \mathrm{Te}$, $\mathrm{H}_2 \mathrm{Se}$, and $\mathrm{H}_2 \mathrm{S}$ would typically have higher boiling points due to increased van der Waals forces (attributable to larger molecular mass and size). But in this case, $\mathrm{H}_2 \mathrm{O}$ surpasses them because hydrogen bonds are much stronger than van der Waals forces.

Statement II states that based on molecular mass alone, $$\mathrm{H}_2 \mathrm{O}$$ should have a lower boiling point than the other Group 16 hydrides but has a higher boiling point due to extensive hydrogen bonding. This statement is also true. Water, having a lower molecular mass, would typically have a lower boiling point if we consider only molecular mass. However, the strong hydrogen bonds between water molecules contribute to a much higher boiling point. Hydrogen bonding significantly impacts physical properties like boiling point because it requires more energy to break these bonds during the phase change from liquid to gas.

In summary, based on the analysis of the two statements:

Option A: Both Statement I and Statement II are true is the correct answer. Each statement is correct, and both relate accurately to the anomalous behavior of water among the hydrides of Group 16 elements.