Given that

Mass of the coin = m

Ten coins are placed on top of each other

a)

The force applied to the seventh coin will be equal to the sum of the weights of the three coins above it.

Number of coins above the seventh coin = 3

Mass of the three coins, M = m + m + m = 3 m

Acceleration due to gravity = g m/s^2

Force on the 7th coin = weight of the coins = Mg = 3mg

b)

Force acting on the eight coin F_8 = mass * g = (m+m) *g = 2mg

The eighth coin exerts force on the seventh coin F_7 = F_8 + \text{weight of 8th coin}

F_7 = 2 mg + mg

F_7 = 3mg

c)

Force acting on the 6th coin F_6 = \text{Force on the 7th coin + weight of the 7th coin }

F_6 = 3mg + mg = 4 mg

According to the newtons third law of motion

The force exerted on the 6th coin due to the 7th coin is the same as the normal reaction force on the 7th coin due to the 6th coin.

Hence, the reaction of 6th coin on the 7th coin = 4mg