Answer
$1.48\;\rm s$
Work Step by Step
First of all, we need to draw the force diagram of the two blocks, as we see below.
We analyzed the weight force of the two blocks into their components
Now we can apply Newton's second law on both blocks.
From the given figure, it is obvious that the total distance traveled down the inclined by block A is 2 meters, so to find the time it takes during this trip we need to find its acceleration.
From the left figure below, we can see that
$$\sum F_{xA}=F_{B\;on\;A}+m_{A}g\sin\theta-f_{kA}=m_{A}a_{xA}$$
where $F_{B\;on\;A}$ and $F_{A\;on\;B}$ are action/reaction pair.
Solving for the acceleration $a_{xA}$;
$$a_{xA}=\dfrac{F_{B\;on\;A}+m_{A}g\sin\theta-f_{kA}}{m_{A}}\tag 1$$
Now we need to find the kintic friction force $f_{kA}$ which is given by
$$f_{kA}=\mu_{kA}F_{nA}\tag 2$$
So that we need to find the normal force on block A by applying Newton's second law in $y$-direction.
$$\sum F_{yA}=F_{nA}-m_Ag\cos\theta=m_{A}a_{yA}=m_{A}(0)=0$$
Thus,
$$F_{nA}=m_Ag\cos\theta $$
Plugging into (2);
$$f_{kA}=\mu_{kA}m_Ag\cos\theta $$
Plugging into (1);
$$a_{xA}=\dfrac{F_{B\;on\;A}+m_{A}g\sin\theta-\mu_{kA}m_Ag\cos\theta}{m_{A}} $$
Since the two blocks are moving down the inclined as one unit, they have the same acceleration. So that $a_{xA}=a_{xB}=a_x $
$$a_{x}=\dfrac{F_{B\;on\;A}+m_{A}g\left[\sin\theta-\mu_{kA} \cos\theta\right]}{m_{A}} \tag 3$$
Now we need to find the force $F_{B\;on\;A}$ which represents the action reaction pair.
Applying Newton's second law in $x$-direction for block B;
$$\sum F_{xB}=m_Bg\sin\theta-F_{A\;on\;B}-f_{kB}=m_Ba_{xB}$$
$$m_Bg\sin\theta-F_{A\;on\;B}-f_{kB}=m_Ba_{x }$$
Solving for $a_{x }$;
$$a_{x }=\dfrac{ m_Bg\sin\theta-F_{A\;on\;B}-f_{kB} }{m_B}\tag 4$$
Now we need to find $f_{kB}$, which is given by
$$f_{kB}=\mu_{kB}F_{nB}\tag 5$$
$$\sum F_{yB}=F_{nB}-m_Bg\cos\theta=m_Ba_{yB}=m_B(0)=0$$
$$F_{nB}=m_Bg\cos\theta$$
Plugging into (5);
$$f_{kB}=\mu_{kB} m_Bg\cos\theta$$
Plugging into (4);
$$a_{x }=\dfrac{ m_Bg\sin\theta-F_{A\;on\;B}-\mu_{kB} m_Bg\cos\theta }{m_B} $$
$$a_{x }=\dfrac{ -F_{A\;on\;B}+m_Bg\left[\sin\theta-\mu_{kB} \cos\theta\right] }{m_B} $$
Solving for $F_{A\;on\;B}$
$$-F_{A\;on\;B}+m_Bg\left[\sin\theta-\mu_{kB} \cos\theta\right] =m_Ba_x$$
$$ F_{A\;on\;B}=m_Bg\left[\sin\theta-\mu_{kB} \cos\theta\right] -m_Ba_x$$
Plugging into (3);
$$a_{x}=\dfrac{m_Bg\left[\sin\theta-\mu_{kB}\cos\theta\right] -m_Ba_x+m_{A}g\left[\sin\theta-\mu_{kA} \cos\theta\right]}{m_{A}} $$
Solving for $a_x$;
$$m_{A}a_{x}+m_Ba_x= m_Bg\left[\sin\theta-\mu_k \cos\theta\right] +m_{A}g\left[\sin\theta-\mu_{kA} \cos\theta\right] $$
$$a_{x} = \dfrac{m_Bg\left[\sin\theta-\mu_{kB} \cos\theta\right] +m_{A}g\left[\sin\theta-\mu_{kA} \cos\theta\right]}{m_{A}+m_B} $$
Plugging the given;
$$a_{x} = \dfrac{10\cdot 9.8\left[\sin20^\circ-0.15 \cos20^\circ\right] +5\cdot 9.8\left[\sin20^\circ-0.2 \cos20^\circ\right]}{5+10} $$
$$a_x=\bf 1.817\;\rm m/s^2$$
Now we need to use the kinematic formula for displacement which is
$$\Delta x_A=v_{ixA}t+\frac{1}{2}a_{xA}t^2$$
to find the time interval of block A during its trip down the inclined surface.
We assume that the system starts from rest, so $v_{ixA}=0\;\rm m/s$.
$$\Delta x_A=0+\frac{1}{2}a_{x }t^2$$
Solving for $t$;
$$t=\sqrt{\dfrac{2\Delta x_A}{a_x}}$$
Plugging the known;
$$t=\sqrt{\dfrac{2\cdot 2}{1.817}}=\color{red}{\bf 1.48}\;\rm s$$
