as per the question the proton is taken from the point of 175 volt to the point of -55 volt.
the work done in case of electric filed is given as
[tex]W=Vab *q[/tex]
where Vab is the potential difference between two points.
Vab =Va-Vb
=175-[-55]volt
= 230 volt
the charge of proton is [tex]q=1.602*10^{-19}[/tex]
hence the work done will be-
W=230 volt ×[tex]1.602*10^{-19} coulomb[/tex]
=368.46[tex]*10^{-19}[/tex] joule [ans]
