At what temperature, the hydrogen molecules will have same kinetic energy as nitrogen molecules at 280 K?

You are right: at 280 K

In every system at thermal equilibrium all molecules interact with each other by means of random collisions, thus transferring kinetic energy among them. Eventually, each molecule reaches the same average kinetic energy (in time). The heavier molecule will have a lower quadratic speed to balance the mean kinetic energy of lighter molecules, which have higher (in the average) quadratic velocity:

##|K_h| = 1/2m_h|v_h^2| = 1/2m_l|v_l^2| =|K_l|##, where ##|K|=1/2m|v^2|## is the average kinetic energy, and ##|v^2|## represents the average of square velocity.

Absolute temperature T is proportional to total kinetic energy.

If the only kind of kinetic energy which is transferred in the interactions is translational energy (as for hydrogen and nitrogen molecules), the proportionality relation is ##|K| = 3/2 k_BT##, where ##k_B## is Boltzmann constant, given by the universal constant of gases, ##R##, divided by Avogadro's constant.

Hence, if the temperature is the same, also the kinetic energies are the same, and vice versa.