Before you jump into this, you have to stand back, look at it, and decide what's going to happen.
You're pulling on some things that rest on a frictionless surface, so they have to start moving. In fact, they're going to accelerate.
Now, I want you to consider for just a second: What would happen if the two blocks were glued together ? Then, you'd be pulling on a single 18-kg block, with 36N of force, and it would accelerate to the right at [ A = F/M ] = 36/18 = 2 m/s² .
The actual situation is no different. The string between them has no weight, and it's tight, and there are no other forces on the blocks, so the whole combination ... both blocks ... is still going to accelerate to the right at 2 m/s² .
OK. Now, let's look at the individual blocks:
The one in the back ... the 12 kg. It's accelerating at 2 m/s² to the right, so the force on it must be [ F = M A ] = 12 x 2 = 24 N to the right. Where does that force come from ? It's the tension in the string between them.
The one in the front ... the 6-kg. It has 36 N pulling it to the right, and 24 N of string tension pulling it to the left. The net force on this block is (26 - 24) = 12N to the right. Its acceleration is [ A = F/M ] = 12/6 = 2m/s² to the right.
Well, whaddaya know ! Both blocks accelerate at the same rate, in the same direction, just as if they were glued together, and the string between them remains tight, with 24N of tension in it.
With this description, you can handle the FBD on your own, I'm sure.
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
