Please tell this.

Dear Student
Case 1 :   
Energy of photon    E=5 eV

Maximum kinetic energy of photoelectrons       K=2 eV

Using     K=E−ϕ                    where ϕ is the work function

∴   2 = 5−ϕ                
⟹ ϕ = 3 eV

Case 2 :    E=6 eV

∴ Maximum kinetic energy of photoelectrons     K = 6 −3 = 3  eV

thus stopping potential of A relative to C is  −3 V.

Regards