To solve this problem, we need to analyze the boiling points of the hydrides of Group 16 elements and the reasons behind them.$\\$Statement I: The boiling point of hydrides of Group 16 elements follows the order $\\$H$_2$O$>$H$_2$Te$>$H$_2$Se$>$H$_2$S$. \\$Statement II: On the basis of molecular mass, H$_2$O is expected to have a lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H$_2$O, it has a higher boiling point.$\\$Let's analyze these statements:$\\$• The boiling points of hydrides generally increase with increasing molecular mass. However, H$_2$O is an exception due to hydrogen bonding.$\\$• H$_2$O has a higher boiling point than other Group 16 hydrides because of the strong hydrogen bonds between water molecules.$\\$• In contrast, H$_2$S, H$_2$Se, and H$_2$Te have weaker van der Waals forces and no hydrogen bonding, leading to lower boiling points.$\\$• Therefore, the order H$_2$O$>$H$_2$Te$>$H$_2$Se$>$H$_2$S is correct.$\\$• Statement II correctly explains why H$_2$O has a higher boiling point despite its lower molecular mass compared to other hydrides.$\\$Conclusion:$\\$Both Statement I and Statement II are true.$\\$Therefore, the correct answer is Option 3.