= Solution
Set the red image at coordinate zero, the blue image at $\delta=0.010\,\mathrm m$, and the screen at $D=1.000\,\mathrm m$. Let the correcting <thin lens> be at $x$, with $\delta<x<D$. Both original images are then real objects for the correcting <thin lens>, with positive object distances $u_r=x$, $u_b=x-\delta$, and common positive image distance $v=D-x$. The <thin lens> equation gives
$$
\boxed{f_r=\frac{x(D-x)}D,\qquad f_b=\frac{(x-\delta)(D-x)}{D-\delta}.}
$$
The <refractive indices> and <Lensmaker's equation> require
$$
\frac{f_b}{f_r}=\frac{n_r-1}{n_b-1}=\frac{10}{11}=\rho.
$$
Substitute the two <focal lengths> and cancel $D-x$:
$$
D(x-\delta)=\rho x(D-\delta),\qquad
x=\frac{D\delta}{D-\rho(D-\delta)}=0.1000\,\mathrm m.
$$
Thus the correcting <thin lens> is $100\,\mathrm{mm}$ beyond the red image and $900\,\mathrm{mm}$ before the screen. The required <focal lengths> and surface radius are
$$
\boxed{f_r=90.0\,\mathrm{mm},\qquad f_b=81.818\ldots\,\mathrm{mm},\qquad R=2(n_r-1)f_r=72.0\,\mathrm{mm}.}
$$
The blue <focal length> is shorter because the blue <refractive index> is larger; these two differing object distances are mapped to the same screen, removing the specified longitudinal <chromatic aberration>.
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