When assembling a ball of charge, why do we need less energy to assemble a bigger ball? The energy required to assembly a ball of charge $ Q $ and radius $ r $ is
$$ U(r) = \frac{3}{5} \frac{Q^2}{4\pi \epsilon_0 r}$$
Why do we need less energy to assemble, let's say, a ball of radius 30m than a ball of radius 5m?
Don't we need to do more work every time we bring a little amount of charge from infinity to the ball and the ball gets more charge? The electric field gets bigger and makes us work harder to bring the following amount of charge.
 A: 
Don't we need to do more work every time we bring a little amount of
charge from infinity to the ball and the ball gets more charge? The
electric field gets bigger and makes us work harder to bring the
following amount of charge.

And why does this apply to a larger ball of charge any more than it would apply to a smaller ball of charge? It applies equally.
You seem to picture a larger ball as having more charge Q but that's ignoring the scenario you actually set up for yourself. Take a closer look at your your equation, the charge Q is the same. It's only the radius that changes.
So in your scenario a larger ball doesn't have more charge. It just has more volume. That means the charges are farther apart and therefore repel less which means less force is required to assemble the charges. But the small ball and large ball still both contain the same amount of charge.
A: It takes less work to assemble a bigger ball if we're considering two balls with the same charge. This should make sense intuitively. If we have have two balls with the same charge but different radii, the charges in the smaller ball are closer together (i.e., the charge density is larger), and it takes more work to get the charges closer to each other.
In the scenario you're thinking about, where we have a ball of a given size and we grow it by adding more charge from infinity, the new, larger ball (with more charge) does cost more energy to assemble than the original, smaller ball.
You probably want to think about the case where we have two balls with the same charge density $\rho$, which is easier to examine if we rewrite your formula in terms of $\rho$. Since
\begin{align}
\rho = \frac{Q}{\frac{4}{3}\pi r^3}
\end{align}
the charge is
\begin{align}
Q = \frac{4}{3}\pi r^3\rho
\end{align}
and the electrostatic energy is
\begin{align}
U &= \frac{3}{5}\frac{kQ^2}{r}\\
&= \frac{3}{5}\frac{k}{r}\left(\frac{16}{9}\pi^2 r^6 \rho^2\right)\\
&= \frac{16}{15}\pi^2k\rho^2 r^5.
\end{align}
So the energy does grow with the size of the ball, in fact quite rapidly as $U \propto r^5$.
