1) First of all, let's calculate the potential difference between the initial point (infinite) and the final point (d=0.529x10-10 m) of the electron.
This is given by:
[tex]\Delta V =- \int\limits^{d}_{\infty} {E} \, dr[/tex]
Where E is the electric field generated by the proton, which is
[tex]E=k_e \frac{q}{r^2} [/tex]
where [tex]k_e=8.99\cdot10^9~Nm^2C^{-2} [/tex] is the Coulomb constant and [tex]q=1.6\cdot10^{-19}~C[/tex] is the proton charge.
Replacing the electric field formula inside the integral, we obtain
[tex]\Delta V =- \int\limits^{d}_{\infty} {k_e \frac{q}{r^2} } \, dr = k_e \frac{q}{d}= 27~V[/tex]
2) Then, we can calculate the work done by the electric field to move the electron (charge [tex]q_e=-1.6\cdot10^{-19}C[/tex]) through this [tex]\Delta V[/tex]. The work is given by
[tex]W=-q_e \Delta V = - (-1.6\cdot10^{-19}C) (27V)=4.35\cdot10^{-18}~J[/tex]
