Mass of each coin $= m$(a) If $F_7$ be the force on 7$^{\text{th}}$ coin (counted from the bottom) experienced due to all coins above it, then $F_7$ = weight of three coins above it = 3 mg N (downwards)(b)$F_{87}$ = force on $7^{\text{th}}$ coin by 8$^{\text{th}}$ coin, then the 8$^{\text{th}}$ coin has to support the weight of the two coins above it. So, the 8th coin shall exert the force $F_{87}$ such that $F_{87}$ = weight of 8th coin + weight of two coins above the 8$^{\text{th}}$ coin $= mg + 2mg= 3mg (N)$ and it acts downwards.(c) The sixth coin experiences force equal to weight of the four coins above it. Hence reaction due to 6$^{\text{th}}$ coin on 7$^{\text{th}}$ coin = $4mg\\ N$ and it acts vertically upwards.