Calculate Total Collisions of Free Electrons in Extension Cord

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To calculate the total number of collisions of free electrons in a 3.00 m copper extension cord carrying a current of 8.0 A, the drift velocity is given as 3.3 x 10^-4 m/s. The user is attempting to relate drift velocity to the time between collisions (Tau) but is struggling to determine the electric field necessary for the calculations. The resistivity of copper and the concentration of free electrons are provided, but the user feels stuck without additional guidance. Suggestions for finding the answer or further resources are requested, highlighting the need for clarity on the electric field's role in the calculations.
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A current of I=8.0\;\rm A is flowing in a typical extension cord of length L=3.00 m. The cord is made of copper wire with diameter d=1.5mm.

The charge of the electron is e=1.6 *10^{-19} C. The resistivity of copper is \rho=1.7 *10^{-8}\; Omega*m. The concentration of free electrons in copper is n=8.5 *10^{28}\;m^{-3}.

Find the total number of collisions ( N_c) that all free electrons in this extension cord undergo in one second.

I know Velocity(drift) = 3.3 * 10^-4 m/s.

I tried to relate the formula V_d = a*Tau, to find Tau the time between collisions...

However given the information I was stuck solving for a = eE/M_e, because I could not determine the electric field...

Can someone please assist me on what information I might be missing?
 
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anyone?
 
im still stuck on this question...
 
ok so since no one seems to have a clue about this problem...

any suggestions of where I may be able to find the answer?
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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