Answer
The electric field is radially outward.
Work Step by Step
The charge of the rod is $3.40\times 10^{-12}~C$
At a radius of $5.00R_1$, only the rod (and not the shell) is enclosed within a cylindrical Gaussian surface.
We can use Equation (23-12) to find the electric field:
$E = \frac{\lambda}{2\pi~\epsilon_0~r}$
$E = \frac{3.40\times 10^{-12}~C/11.00~m}{(2\pi)~(8.854\times 10^{-12}~F/m)~(0.00650~m)}$
$E = 0.855~N/C$
The positive sign of $E$ shows that the electric field is radially outward.
