The temperature at which the kinetic energy of oxygen molecules...

The kinetic energy is given by $KE = \dfrac{f}{2}kT$.

Let the kinetic energy be ${K}_{1}$ at $27 ° C$ and the kinetic energy be $K '$ at $T '$ temperature. As given in the question, $K ' = 2{K}_{1}$. Hence,

$\dfrac{K '}{{K}_{1}} = \dfrac{T '}{T} ⇒ 2 = \dfrac{T '}{300} ⇒ T ' = 600K$

In Celsius the temperature is $T ' - 273 = 327 ° C$.