here it behaves like two capacitors connected in parallel.
let a = 2 mm, b = 6 mm, c = 7 mm
C1/L = 2*pi*epsilon/ln(b/a)
= 2*pi*8.854*10^-12/ln(6/2)
= 5.06*10^-11 F/m
C2/L = 2*pi*epsilon/ln(c/b)
= 2*pi*8.854*10^-12/ln(7/6)
= 3.61*10^-10 F/m
Ceq/L = C1/L + C2/L
= 5.06*10^-11 + 3.61*10^-10
= 4.12*10^-10 F/m
= 412*10^-12 F/m
= 412 pF/m <<<<<<<--------------Answer
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