A. Lithium
The equation for the photoelectric effect is:

where
is the energy of the incident light, with h being the Planck constant, c being the speed of light, and
being the wavelength
is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)
K is the maximum kinetic energy of the photoelectron
In this problem, we have
, so the energy of the incident light is

Converting in electronvolts,

Since the electrons are emitted from the surface with a maximum kinetic energy of
K = 4.0 eV
The work function of this metal is

So, the metal is Lithium.
B. cesium, potassium, sodium
The wavelength of green light is

So its energy is

Converting in electronvolts,

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:
Cesium
Potassium
Sodium
C. 4.9 eV
In this case, we have
- Copper work function: 
- Maximum kinetic energy of the emitted electrons: K = 2.7 eV
So, the energy of the incident light is

Then the copper is replaced with sodium, which has work function of

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be
