Correct Answer - C
Here length L = 20 cm `= 20 xx 10^(-2)m`
inner radius `r_(1) = 1.5 cm = 1.5 xx 10^(-2)m`
outer radius `r_(o) = 1.6 cm = 1.6 xx 10^(-2)m`
charge `q = 4 muC = 4 xx 10^(-6)C`
Capacitance ,
`C = (2pi epsi_(0)L)/(log_(e)((r_(o))/(r_(i)))) = (2pi epsi_(0)L)/(2.3 log_(10)((r_(o))/(r_(i))))= (2pi xx 8.85 xx 10^(-12) xx 20 xx 10^(-2))/(2.3 log_(10)((1.6 xx 10^(-2))/(1.5 xx 10^(-2)))) `
` = 1.7 xx 10^(-10)` F