To find the electric field at a distance $R$ (where $a < R < b$), we use Gauss's Law. We construct a spherical Gaussian surface of radius $R$ concentric with the system.

According to Gauss's Law: $\oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{enclosed}}{\varepsilon_0}$.

The charge enclosed ($q_{enclosed}$) by this surface is the total charge on the inner sphere, which is $+3Q$. The outer shell's charge ($-Q$) lies outside the Gaussian surface and does not contribute to the electric flux through it (the field inside a uniformly charged spherical shell due to the shell itself is zero).

Thus, $E (4\pi R^2) = \frac{3Q}{\varepsilon_0}$. Solving for $E$, we get $E = \frac{3Q}{4\pi\varepsilon_0 R^2}$.