Charge density!

Consider a very long section of the rod and shell, of length $L >> R$ so that boundary effects are negligible.

Choose a Gaussian surface: $\mathcal S$ inside the the conducting shell. Since the field in a conductor is zero, the total flux through $\mathcal S$ is $\Phi = 0$ , so the enclosed charge is also zero.

This is the sum of the negative surface charge $-A\sigma$ on the inside of the shell and of the positive charge $L\lambda$ of the rod: $A\sigma + L\lambda = 0$ . Since the net charge on the shell is supposedly zero, the positive surface charge on the outside is $+A\sigma$ .

(In other words, the outside surface charge on the shell is equal to the charge of the rod inside .)

We obtain $\sigma = \frac{L\lambda}A = \frac{L\lambda}{2\pi R L} = \frac{\lambda}{2\pi R} = \frac{\SI{5}{nC/m}}{2\pi\cdot \SI{0.03}{m}} = \boxed{\SI{26.53}{nC/m^2}}.$