May be this is easier than you think

In general, for a charge distribution of characteristic size $a$ and total charge $q$ , $E \propto \frac q{a^2}.$ The cube that is cut out has characteristic size $a' = \tfrac12 a$ and charge $q' = \tfrac18 q$ . Thus $\frac{E'}E = \frac{q'/a'^2}{q/a^2} = \frac{q'}{q}\cdot \left(\frac{a}{a'}\right) = \tfrac18(2)^2 = \tfrac12$ . Removing this cube, then, reduces the electric field to $E_2 = E - E' = E - \tfrac12 E = \tfrac12E,$ so that the desired ratio is $\boxed{1/2}$ .