Correct Answer - D
As is clear from Fig 40, three rows of capacity `C_(1) , (C_(1))/(2)` and `C_(1)` are connected in parallel between the points `A` and `B`.
Total capacity, `C = C_(1) + (C_(1))/(2) + C_(1) = (5)/(2) C_(1)`
As `q = CV :. 1.5xx10^(-6) = (5)/(2) C_(1) xx 6`
`C_(1) = (3)/(30) xx 10^(-6) F = 0.1 mu F`