A metal surface has a work function of 2.4 eV…

Step 1. (a) $\lambda_0 = \dfrac{1240\ \text{eV nm}}{2.4\ \text{eV}} = 516.7\ \text{nm}$, green light.

Step 2. (b) $\dfrac{1240\ \text{eV nm}}{405\ \text{nm}} = 3.06\ \text{eV}$, so $KE_{\max} = 3.06\ \text{eV} - 2.4\ \text{eV} = 0.66\ \text{eV}$.

Step 3. (c) $0.66\ \text{eV} = 1.06 \times 10^{-19}\ \text{J}$, so $v = \sqrt{\dfrac{2 \times 1.06 \times 10^{-19}\ \text{J}}{9.1 \times 10^{-31}\ \text{kg}}} = 4.82 \times 10^{5}\ \text{m/s}$.