As shown in the figure, two blocks are in contact sliding down an...

Let us first calculate the acceleration of each block as if they were sliding independently.

For the $2.0 \text{ kg}$ block (upper block):
a_1 = g(\sin 30^{}^{{}^{\circ}} - \mu_1 \cos 30^{}^{{}^{\circ}})
$a_1 = 10 \left(0.5 - 0.20 \times \dfrac{\sqrt{3}}{2}\right) = 5 - \sqrt{3} \approx 3.27 \text{ m/s}^2$

For the $4.0 \text{ kg}$ block (lower block):
a_2 = g(\sin 30^{}^{{}^{\circ}} - \mu_2 \cos 30^{}^{{}^{\circ}})
$a_2 = 10 \left(0.5 - 0.30 \times \dfrac{\sqrt{3}}{2}\right) = 5 - 1.5\sqrt{3} \approx 2.40 \text{ m/s}^2$

Since $a_1 > a_2$, the upper $2.0 \text{ kg}$ block will tend to slide faster than the lower $4.0 \text{ kg}$ block. Because it is behind the $4.0 \text{ kg}$ block, it will push against it, and both blocks will move together with a common acceleration $a$.

Considering both blocks as a single system of mass $M = m_1 + m_2 = 2.0 + 4.0 = 6.0 \text{ kg}$:
The equation of motion along the incline is:
(m_1 + m_2)a = (m_1 + m_2)g \sin 30^{}^{{}^{\circ}} - f_1 - f_2
where $f_1$ and $f_2$ are the kinetic friction forces on the two blocks.

f_1 = \mu_1 m_1 g \cos 30^{}^{{}^{\circ}} = 0.20 \times 2.0 \times 10 \times \dfrac{\sqrt{3}}{2} = 2\sqrt{3} \text{ N}
f_2 = \mu_2 m_2 g \cos 30^{}^{{}^{\circ}} = 0.30 \times 4.0 \times 10 \times \dfrac{\sqrt{3}}{2} = 6\sqrt{3} \text{ N}

Substituting these into the equation of motion:
$6.0 \times a = 6.0 \times 10 \times 0.5 - (2\sqrt{3} + 6\sqrt{3})$
$6a = 30 - 8\sqrt{3}$
$a = 5 - \dfrac{4\sqrt{3}}{3}$

Using $\sqrt{3} \approx 1.732$:
$a = 5 - \dfrac{4 \times 1.732}{3} = 5 - 2.309 = 2.691 \text{ m/s}^2$

Rounding to one decimal place, we get $a \approx 2.7 \text{ m/s}^2$.