Answer
$\therefore \text d) \lambda _2\gt \lambda _1\gt \lambda _3$ is the right option
Work Step by Step
$\lambda _1$, $\lambda _2$, $\lambda _3$ are the three wavelengths of light
$$\lambda \propto \frac{1}{Energy(E)}$$
Since,
Kinetic Energies of the three lights are as Follows
$\lambda _1$ : $2.9\times 10^{-20}J$
$\lambda _2$ : ${0}J$
$\lambda _1$ : $4.2\times 10^{-19}J$
$\therefore \lambda _2\gt \lambda _1\gt \lambda _3$
