Given: Refer the attachment diagram for data

m₁ = 100 g = 0.1 kg , m₂ = 500 g = 0.5 kg , m₃ = 50 g = 0.05 kg

The free-body diagram for the system is given in attachment.

From the free-body diagram of the 500 g block,

T + 0.5a − 0.5g = 0 …..(i)

From the free-body diagram of the 50 g block,

T₁ + 0.05g − 0.05a = a ….(ii)

From the free-body diagram of the 100 g block,

T₁ + 0.1a − T + 0.5g = 0 ….(iii)

From equation (ii), T₁ = 0.05g + 0.05a …..(iv)

From equation (i), T₁ = 0.5g − 0.5a …..(v)

Equation (iii) becomes T₁ + 0.1a − T + 0.05g = 0

From equations (iv) and (v), we get:

0.05g + 0.05a + 0.1a − 0.5g + 0.5a + 0.05g = 0

0.65a = 0.4 g

⇒ a=g = g= g downward

So, the acceleration of the 500 gm block is (downward).