What percentage of the projectile's incident kinetic energy does the target carry off after being struck?

I think we want to start with conservation of linear momentum.

m?v?? + m?v?? = m?v?? + m?v??

Since the 2 objects stick together and travel together  v?? = v?? so we can just call this v? and factor it out.

Also v?? = 0m/s so we can eliminate that term.

m?v??= (m? + m?)v?

We can use this to find the final velocity.

v? = m?v??/(m? + m?) = [(0.24kg)/(0.24kg+2.74kg)]v?? ? .08v??

Now let's look at the projectile's change in (kinetic) energy.  We want to know by how much it was reduced so let's look at a ratio of the final kinetic energy to the initial kinetic energy.

KEf/KEi = (½mv?²)/(½mv?²) = [½m(.08v?)²]/(½mv?²) = .0064

This says that the porjectile has 0.64% of the kinetic energy it started with.

The target took 99.36%