A diatomic gas consisting of rigid molecules is at a temperature of...

Recall the formula of the kinetic energy of one molecule in terms of Boltzmann constant and absolute temperature,

$K . E . = \dfrac{1}{2}KT$, for two rigid molecules, it will be,

$K . E . = 2 \times \dfrac{1}{2}KT = KT$, if~$I$~is~a moment of inertia of body and~$\omega$~is the angular velocity then,

$KT = \dfrac{1}{2}I{ω}^{2} ⇒ \omega = \sqrt{\dfrac{2KT}{I}} = \sqrt{\dfrac{2 \times 1 . 38 \times {10}^{- 23}\left(87 + 273\right)}{2 . 76 \times {10}^{- 39} \times {10}^{- 7}}} = 6 \times {10}^{12}rads^{- 1}$