= Solution
Use the standard infinite-plane boundary condition: the sheet is an <equipotential>, with no imposed uniform background <electric field> and the <electric potential> tending to zero far from the localized charges. The images of $+Q$ at $z=d$ and $-2Q$ at $z=2d$ are $-Q$ at $z=-d$ and $+2Q$ at $z=-2d$.
At the point $z=0^+$ directly below the real charges, horizontal components vanish. The first real-image pair contributes $-2k_eQ/d^2$ along $z$, while the second contributes $+2k_e(2Q)/(2d)^2=+k_eQ/d^2$. Therefore
$$
\boxed{\mathbf E(0,0,0^+)=-\frac{Q}{4\pi\epsilon_0d^2}\hat{\mathbf z}}.
$$
For $Q>0$, the <electric field> points downwards into the sheet. The expression is the one-sided vacuum limit; the <electric field> inside the conducting material is zero. A prescribed uniform background field would be an additional boundary condition and would change this result.
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