#### Answer

(a) Please refer to the free body diagram below.
(b) The ground exerts a force of 0.0493 N on the froghopper.
(c) The normal force is 409 times the froghopper's weight.

#### Work Step by Step

(a) Please refer to the free body diagram below.
(b) We can find the acceleration during the jump.
$a = \frac{v-v_0}{t} = \frac{4.0~m/s - 0}{0.0010~s}$
$a = 4000~m/s^2$
We can use a force equation to find $F_N$.
$\sum F = ma$
$F_N - mg = ma$
$F_N = ma + mg$
$F_N = (1.23\times 10^{-5}~kg)(4000~m/s^2)+(1.23\times 10^{-5}~kg)(9.80~m/s^2)$
$F_N = 0.0493~N$
The ground exerts a force of 0.0493 N on the froghopper.
(c) We can express the force in terms of the froghopper's weight $mg$.
$F = \frac{0.0493~N}{(1.23\times 10^{-5}~kg)(9.80~m/s^2)}$
$F = 409\times ~weight$
The normal force is 409 times the froghopper's weight.