Block 1 (the rightmost block) has
• net horizontal force
∑ <em>F</em> = <em>F</em> - <em>T₁</em> - <em>f₁</em> = <em>m₁a</em>
• net vertical force
∑ <em>F</em> = <em>N₁</em> - <em>m₁g</em> = 0
where <em>F</em> = 57.3 N, <em>T₁</em> is the tension in the string connecting blocks 1 and 2, <em>f₁</em> is the magnitude of kinetic friction felt by block 1, <em>m₁</em> = 0.8 kg is its mass, <em>a</em> is the acceleration you want to find, and <em>N₁</em> is the magnitude of the normal force exerted by the surface.
Block 2 (middle) has much the same information:
• net horiz. force
∑ <em>F</em> = <em>T₁</em> - <em>T₂</em> - <em>f₂</em> = <em>m₂a</em>
• net vert. force
∑ <em>F</em> = <em>N₂</em> - <em>m₂g</em> = 0
with similarly defined symbols.
The same goes for block 3 (leftmost):
• net horiz. force
∑ <em>F</em> = <em>T₂</em> - <em>f₃</em> = <em>m₃a</em>
• net vert. force
∑ <em>F</em> = <em>N₃</em> - <em>m₃g</em> = 0
We have <em>m₁</em> = <em>m₂</em> = <em>m₃</em> = 0.8 kg, so I'll just replace each with <em>m</em>. It follows that each normal force has the same magnitude, <em>N₁</em> = <em>N₂</em> = <em>N₃</em> = <em>mg</em>. And as a consequence of that, each frictional force has the same magnitude, <em>f₁</em> = <em>f₂</em> = <em>f₃</em> = 0.4<em>mg.</em>
In short, the relevant equations are
[1] … 57.3 N - <em>T₁</em> - 0.4<em>mg</em> = <em>ma</em>
[2] …<em>T₁</em> - <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>
[3] … <em>T₂</em> - 0.4<em>mg</em> = <em>ma</em>
<em />
Adding [1], [2] and [3] together eliminates the tension forces, and we get
57.3 N - 1.2<em>mg</em> = 3<em>ma</em>
<em />
Solve for <em>a</em> :
57.3 N - 1.2 (0.8 kg) (9.8 m/s²) = 3 (0.8 kg) <em>a</em>
57.3 N - 9.408 N = (2.4 kg) <em>a</em>
<em>a</em> = (47.892 N) / (2.4 kg)
<em>a</em> ≈ 20.0 m/s²
