= Solution
Take the grounded <electrical conductor> to occupy $z\leq0$ and approximate the negligible-radius charged sphere by a <point charge> $Q$ at $\mathbf r_+=D\widehat{\mathbf z}$. Put a fictitious <image charge> $-Q$ at $\mathbf r_-=-D\widehat{\mathbf z}$. The constructed <electric potential> is zero on $z=0$, has the correct <point charge> singularity, and vanishes at infinity. The <uniqueness theorem for the electrostatic potential> makes it the physical <electric potential> for $z>0$:
$$
\phi(\mathbf r)=\frac Q{4\pi\epsilon_0}
\left(\frac1{|\mathbf r-D\widehat{\mathbf z}|}
-\frac1{|\mathbf r+D\widehat{\mathbf z}|}\right).
$$
Taking its negative <gradient> gives the requested <electric field> outside the small sphere:
$$
\boxed{\mathbf E(\mathbf r)=\frac Q{4\pi\epsilon_0}
\left[\frac{\mathbf r-D\widehat{\mathbf z}}{|\mathbf r-D\widehat{\mathbf z}|^3}
-\frac{\mathbf r+D\widehat{\mathbf z}}{|\mathbf r+D\widehat{\mathbf z}|^3}\right],\quad z>0.}
$$
On the plane at cylindrical <radius> $r$, the tangential components cancel and the normal component is
$$
E_z(r,0^+)=-\frac{QD}{2\pi\epsilon_0(r^2+D^2)^{3/2}}.
$$
The <electrostatic boundary conditions at a conductor>, with outward <unit normal> $+\widehat{\mathbf z}$ and zero interior <electric field>, give the induced <surface charge density>
$$
\boxed{\sigma(r)=\epsilon_0 E_z(r,0^+)
=-\frac{QD}{2\pi(r^2+D^2)^{3/2}}.}
$$
Finally integrate over annuli of <area> $2\pi r\,dr$:
$$
Q_{\rm ind}=\int_0^\infty\sigma(r)2\pi r\,dr
=-QD\int_0^\infty\frac{r\,dr}{(r^2+D^2)^{3/2}}
=-QD\left[-\frac1{\sqrt{r^2+D^2}}\right]_0^\infty,
\qquad\boxed{Q_{\rm ind}=-Q.}
$$
The <image charge> is a computational device, not a real charge below the plane. For a finite conducting sphere, its induced polarization would change the point-source approximation; the stated negligible <radius> is essential.
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