All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass $$2 \text{ kg}$$ is

You have:

• a 2 kg block on a 30° incline
• connected by a string over a pulley
• then a movable pulley holding a 4 kg mass

Key idea: movable pulley changes both force and motion relation

Step 1: understand tension

Let tension in string = T

For the 4 kg block:
It is supported by two segments of the string
So upward force = 2T

Equation:
$$2T−4g=4a₄.$$

Step 2: relation of acceleration

Very important part

If the 2 kg block moves up the incline by x
the movable pulley (and 4 kg mass) moves down by x/2

So:
$$a₂=2a₄$$

Now lets write equation for the 2 Kg block

$$m_2g\sin\ \theta\ -T=m_2a_2$$

 $$2g\times\ \frac{1}{2}-T=2a_2$$

$$g-T=4a_4$$

$$2T−4g=4a₄.$$

now on multiply and adding both so that T gets cancelled

$$-2g=12a_4$$

$$\ a_4=-\frac{g}{6}\left(negative\ as\ we\ considered\ opposite\ direction\right)$$

so now we know 

$$a₂=2a₄$$

so 

$$a₂=2\times\ \frac{g}{6}$$

$$a₂=\ \frac{g}{3}$$