A copper wire is in the form of a cylinder and has a resistance R. It is stretched till its thickness reduces by half of its initial size. Find its new resistance in terms of R

If you stretch the wire till it's radius is halved the previous value, then it's length will increase 4 times to keep the volume constant. 

Initial volume = LA = Lπr²
Since now thickness is reduced by half so r' = r/2

New volume = L'A' = L'πr'² = L'π(r/2)² = L'πr²/4 = L'A/4

Since volume should be the same so L'A' = L'A/4 = LA

So L' = 4L

So, new values are, 
L' = 4L
A' = A/4 = (π*(r/2)²) 
Where, L and A are the values of original length and area. 
Resistivity remains the same as it depends upon the material of the wire. 
So, the new resistance will be, 
R = ρL'/A' = ρ(4L)/(A/4) = 16ρL/A
    = 16r
Where, r is the value of original resistance.

Hope this helps.
​​​​​​​Regards