Answer:
(a).The maximum current in the wire is
.
(b). The electric field in the wire is
.
(c).The current also decrease with time.
(d). The total amount of energy dissipated in the wire is 
Explanation:
Given that,
Diameter of metal plates = 10 cm
Distance between the plates = 1.0 cm
Charged = 12.5 nC
Diameter of copper wire = 0.224 mm
We need to calculate the cross section area of the plates
Using formula of area

Put the value into the formula


We need to calculate the capacitor
Using formula of capacitor

Put the value into the formula


We need to calculate the resistance of the wire
Using formula of resistivity

Put the value into the formula


We need to calculate the voltage
Using formula of charge


Put the value into the formula


(a). We need to calculate the current
Using formula of current




(b). We need to calculate the electric field
Using formula of electric field

Put the value into the formula


The electric field in the wire is
.
(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.
The current is directly proportional to the voltage between the plates .
Hence, The current also decrease with time.
(d). We need to calculate the total amount of energy dissipated in the wire
Using formula of energy

Put the value into the formula


The total amount of energy dissipated in the wire is 
Hence, (a).The maximum current in the wire is
.
(b). The electric field in the wire is
.
(c).The current also decrease with time.
(d). The total amount of energy dissipated in the wire is 