Temperature at which r.m.s. speed of O₂ is equal to that of neon at 300 K is :

Root mean square velocity = $\sqrt{\frac{3RT}{M}}$
It means it is directly proportional to Temperature and inversely proportional to Molar mass.

$$\begin{gathered} {\text{Root mean square velocity of O}}_{\text{2 }}\text{= }\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}(\mathrm{molar} \mathrm{mass} \mathrm{of} {\mathrm{O}}_{\text{2 }})}} \\ \text{RMS\hspace{0.17em}}{\mathrm{O}}_{\text{2 }} =\sqrt{\frac{3\mathrm{RT}}{\mathrm{molar} \mathrm{mass} \mathrm{of} {\mathrm{O}}_{\text{2 }}}} ----\left(1\right) \\ \text{RMS neon} =\sqrt{\frac{3\mathrm{RT}}{\mathrm{molar} \mathrm{mass} \mathrm{of} \text{neon}}} \\ \mathrm{T} \mathrm{for} \mathrm{neon} = 300\mathrm{K} \\ \mathrm{Mass} \mathrm{of} \mathrm{neon} = 20.1797 \mathrm{g}/\mathrm{mol} \\ \text{RMS neon} =\sqrt{\frac{3\times 300\times \mathrm{R}}{ 20.1797}} -----\left(2\right) \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} , \text{RMS\hspace{0.17em}}{\mathrm{O}}_{\text{2 }} =\text{RMS neon} \\ \mathrm{So}, \mathrm{from} 1 \mathrm{and} 2 \\ \sqrt{\frac{3\mathrm{RT}}{\mathrm{molar} \mathrm{mass} \mathrm{of} {\mathrm{O}}_{\text{2 }}}} =\sqrt{\frac{3\times 300\times \mathrm{R}}{ 20.1797}} \\ \mathrm{squaring} \mathrm{both} \mathrm{side}, \\ \frac{3\mathrm{RT}}{32} =\frac{3\times 300\times \mathrm{R}}{ 20.1797} \\ \frac{3\mathrm{T}}{32}=\frac{900}{20.17} \\ \mathrm{T} =\frac{900\times 32}{20.17\times 3}= 475.9\mathrm{K} \end{gathered}$$
Temperature at which r.m.s. speed of O₂ is equal to that of neon at 300 K is 475.9K

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