Two gases A and B are at absolute temperatures 350 K and 420 K respectively. The ratio of average kinetic energy of the molecules of gas $B$ to that of gas $A$ is

The average kinetic energy for an ideal gas is given by $KE_{\text{avg}} = \frac{3}{2}kT$, where $T$ is the absolute temperature. For gases A and B with temperatures $T_A = 350\text{ K}$ and $T_B = 420\text{ K}$, the ratio of kinetic energies simplifies to: $\frac{KE_{\text{avg,B}}}{KE_{\text{avg,A}}} = \frac{T_B}{T_A} = \frac{420}{350}$ Reducing this fraction yields: $\frac{420}{350} = \frac{42}{35} = \frac{6}{5}$ The ratio of average kinetic energy of gas B to gas A is therefore $6:5$. Final answer: $\boxed{\text{A}}$