The expression that you wrote for the density of states is for a free electron gas in 2D. Your answer is right, if you have a 2D free electron gas. However, for a 3D gas, the density of states per volume is given by $$D(k)=\frac{k^2 }{ \pi^2}$$ If
You get the above expression as follows:
$$D(k)\ dk\times Volume= \underbrace{2}_{\mbox{spin degeneracy}}\times \underbrace{4\pi k^2 dk}_{\mbox{3D volume element in k-space}} \div \underbrace{\left(\frac{2 \pi}{L}\right)^3}_\mbox{Volume occupied by 1 state in k-space}$$
where L is the length of the container.
Some brief explanation of the origins of the terms:
We multiply by 2 for spins because electrons are spin-1/2 particles and therefore each k-state will have a degeneracy of $2(1/2)+1=2$.
The $4 \pi k^2$ factor is obtained following the same reasoning as you would to find the volume occupied between $r$ and $(r+dr)$, by a sphere.
From periodic boundary conditions, we find that states are uniformly spaced by $\frac{2 \pi}{L}$ in k-space. Therefore the volume occupied by each k-state is simply the cube of that.
If you use this expression, you will indeed get the mean energy to be $\frac{3E_F}{5}$.