IMPOSSIBLE PHYSICS PROBLEM!

where: μk1 = coefficient of kinetic friction on Package A (0.220). Substitute into eq. (1):
(3) m1gsin(20°) + Fb - μk1*m1gcos(20°) = m1a

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PACKAGE B:
The same procedure applies to Package B:
ΣFx = max
m2gsin(20°) - Fa - Ff_B = m2a
where: Fa = Force of Package A on Package B (an opposing reaction force since Package B applies a force on Package A)
(4) m2gsin(20°) - Fa - μk2*m2gcos(20°) = m2a

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SOLVE:
(3) m1gsin(20°) + Fb - μk1*m1gcos(20°) = m1a
(4) m2gsin(20°) - Fa - μk2*m2gcos(20°) = m2a

Add equations (3) and (4) together. Note that Fb = Fa. These are two opposing reaction forces that MUST be equal for both boxes to slide down together.
m1gsin(20°) + m2gsin(20°) - μk1*m1gcos(20°) - μk2*m2gcos(20°) = m1a + m2a
gsin(20°)(m1 + m2) + gcos(20°)(-μk1*m1 - μk2*m2) = a (m1 + m2)

a = [gsin(20°)(m1 + m2) + gcos(20°)(-μk1*m1 - μk2*m2)] / (m1 + m2)
a = [9.81*sin(20°)(6+8) + 9.81*cos(20°)(-0.220*6 - 0.170*8)] / (6 + 8)
a = 1.59 m/s^2

This is the acceleration of both blocks. Now, we can find the time it takes package A to reach the bottom:

x = vix*t + 0.5*ax*t^2
where:
x = distance (2 m)
vix = initial velocity of Package A (0 m/s,assumed to start from rest)
ax = 1.59 m/s^2
t = time it takes for the package to reach the bottom

2 = 0.5*1.59*t^2
==>t = 1.586 seconds