Assuming that Artemis and Apollo are of equal mass, so their barycenter is exactly halfway between them:
The umbra (region of total eclipse) forms a cone which can be extended to lie tangent to the surface of the sun, and whose sides are also tangent to the surface of Artemis. Between the center of the sun and the center of Artemis, we have a length of 221,335,690 km. So, we've got a line tangent to the surface of a sphere of radius 0.82 million km at one end and to the surface of a sphere of radius 7806 km at the other end, and we need to figure out where and at what angle that double-tangent line ends up intersecting the line between the sun and Artemis. Recognizing that the tangent and trans-central lines form a pair of similar right triangles with the radii of Artemis and the sun, we can figure that out with the following formula:
$\sin\theta = \frac{r}{x} = \frac{R}{x + D}$
Where $r$ is the radius of Artemis, $R$ is the radius of the sun, $x$ is the distance from the center of Artemis to the apex of the umbral cone, and $D$ is the distance from the center of Artemis to the center of the sun. Rearranging to solve fox $x$, we get:
$x = Dr/(R-r)$
Which, plugging in all our known values, gives us a distance of $x = 2,127,258.26 km$. That's larger than the distance between the planets, so that's good! There will in fact be an area of total eclipse. And the angle ends up being $\theta = 0.00366952037 rad$
Now, we need to figure out where the umbral cone intersects the surface of Apollo. We could do complicated exact intersection calculations, but with such a shallow cone angle (only about 0.21 degrees), it is reasonable to treat the intersection of the umbral cone with Apollo as a cylinder (i.e., having constant radius across the relatively small diameter of Apollo, which we shall assume is much smaller than the distance between the planets), which simplifies the calculations enormously (especially since we don't actually have a figure for the radius or diameter of Apollo).
The distance from the apex of the umbral cone to Apollo will be $2,127,258.26 km - 128,620 km = 1,998,638.26 km$. At that distance, the cone will have a radius of $1,998,638.26 km \sin{\theta} = 7,334.02735 km$
That is considerably larger than the radius of Earth, so if Apollo is approximately Earth-sized, it will be put in complete darkness by Artemis's shadow. If it is larger than that, the proportion of Apollo's surface which sees a total eclipse will depend on figuring out what radius-dependent arc length along the surface corresponds to a chord of that half-length.
The penumbra will be considerably larger, so I feel reasonably confident in stating that a partial eclipse will be visible everywhere on Apollo for any reasonable range of sizes, without doing and specific calculations.
