Calculate the speed at which the bullet leaves the rifle using the principle of conservation of momentum.

Let $m_b$ be the mass of the bullet, $m_r$ be the mass of the rifle.
Let $u_b$ and $u_r$ be their initial velocities, and $v_b$ and $v_r$ be their final velocities.

Given:
$m_b = 20 \text{ g} = 0.020 \text{ kg}$
$m_r = 3 \text{ kg}$
$u_b = 0 \text{ m/s}$ (rifle is stationary)
$u_r = 0 \text{ m/s}$ (rifle is stationary)
$v_r = -1.4 \text{ m/s}$ (recoils, so opposite direction to bullet)

According to the conservation of momentum:

$$m_b u_b + m_r u_r = m_b v_b + m_r v_r$$ $$(0.020 \text{ kg})(0 \text{ m/s}) + (3 \text{ kg})(0 \text{ m/s}) = (0.020 \text{ kg}) v_b + (3 \text{ kg})(-1.4 \text{ m/s})$$ $$0 = 0.020 v_b - 4.2$$ $$0.020 v_b = 4.2$$ $$v_b = \frac{4.2}{0.020}$$ $$v_b = 210 \text{ m/s}$$

The speed at which the bullet leaves the rifle is the magnitude of $v_b$.
The speed is $\boxed{\text{}210 \text{ m/s}\text{}}$

The bullet enters the block with the speed calculated in Step 1 and comes to rest.
Given:
Mass of bullet $m_b = 0.020 \text{ kg}$
Initial velocity of bullet $u = 210 \text{ m/s}$
Final velocity of bullet $v = 0 \text{ m/s}$
Distance travelled $s = 0.4 \text{ m}$

First, calculate the acceleration ($a$) of the bullet using a kinematic equation:

$$v^2 = u^2 + 2as$$ $$(0 \text{ m/s})^2 = (210 \text{ m/s})^2 + 2a(0.4 \text{ m})$$ $$0 = 44100 + 0.8a$$ $$0.8a = -44100$$ $$a = \frac{-44100}{0.8}$$ $$a = -55125 \text{ m/s}^2$$

The negative sign indicates deceleration.

Next, calculate the average force ($F$) using Newton's Second Law:

$$F = m_b a$$ $$F = (0.020 \text{ kg})(-55125 \text{ m/s}^2)$$ $$F = -1102.5 \text{ N}$$

The magnitude of the average force is the absolute value of $F$.
The magnitude of the average force exerted by the block on the bullet is $\boxed{\text{}1102.5 \text{ N}\text{}}$

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. The force exerted by the block on the bullet and the force exerted by the bullet on the block are an action-reaction pair. Therefore, their magnitudes are equal.

The magnitude of the force calculated in QUESTION 4.2 compared to the magnitude of the force exerted by the bullet on the block is $\boxed{\text{}THE SAME\text{}}$