As John Holtz mentions in the comments, the true $h$ value for a planet may be smaller than the value shown in my table. That $h$ value only occurs if the planet's argument of periapsis is ±90°. Fortunately, Neptune's argument of periapsis is currently ~272°, so my $h$ value should be fairly close to the true value.

As John Holtz mentions in the comments, the true $h$ value for a planet may be smaller than the value shown in my table. The table's $h$ value only occurs if the planet's argument of periapsis is ±90°. Fortunately, Neptune's argument of periapsis is currently ~272°, so my $h$ value should be fairly close to the true value.

James K has supplied a list of orbit inclinations to the Solar System's invariable plane. Here's the table using those values.

Planet distance from the Solar System invariable plane.

That reduces Neptune's $h$ considerably! Uranus may even be the current "winner", depending on their arguments of periapsis with respect to the invariable plane.

The thickness of the planetary disc is dominated by Neptune, due to its large orbital radius.

We can calculate a planet's maximum distance from the ecliptic $h$ from the inclination angle of its orbit $\theta$ and its aphelion distance $r$. We get a right triangle, with $r$ as the hypotenuse, so $$h = r\sin\theta$$

The table below was calculated using data from the NASA Planetary Fact Sheet. Angles are in degrees, distances are in millions of kilometres.

Planet distance from the ecliptic plane.

So the total thickness of the disc is $2×142.784 = 285.568$ million kilometres, which is almost $1.91$ au.

Here's the Python code I used to create that table:

from math import sin, radians

names = (
 'Mercury', 'Venus', 'Earth',
 'Mars', 'Jupiter', 'Saturn',
 'Uranus', 'Neptune',
)

# Orbit data from https://nssdc.gsfc.nasa.gov/planetary/factsheet/
# Inclination to ecliptic plane
inc = [7.0, 3.4, 0.0, 1.9, 1.3, 2.5, 0.8, 1.8]
# Aphelion in millions of kilometres
aph = [69.8, 108.9, 152.1, 249.2, 816.6, 1514.5, 3003.6, 4545.7]

print("|Name | Inclination | Aphelion | Distance|")
print("|-|-|-|-|")
for n, th, r in zip(names, inc, aph):
    # Perpendicular distance to eciptic
    h = r * sin(radians(th))
    print(f"|{n} | {th} | {r} | {h:.3f}|")

Here's a live version of the script running on the SageMathCell server.

The thickness of the planetary disc is dominated by Neptune, due to its large orbital radius.

We can calculate a planet's maximum distance from the ecliptic $h$ from the inclination angle of its orbit $\theta$ and its aphelion distance $r$. We get a right triangle, with $r$ as the hypotenuse, so $$h = r\sin\theta$$

The table below was calculated using data from the NASA Planetary Fact Sheet. Angles are in degrees, distances are in millions of kilometres.

Planet distance from the ecliptic plane.

So the total thickness of the disc is $2×142.784 = 285.568$ million kilometres, which is almost $1.91$ au.

Here's the Python code I used to create that table:

from math import sin, radians

names = (
 'Mercury', 'Venus', 'Earth',
 'Mars', 'Jupiter', 'Saturn',
 'Uranus', 'Neptune',
)

# Orbit data from https://nssdc.gsfc.nasa.gov/planetary/factsheet/
# Inclination to ecliptic plane
inc = [7.0, 3.4, 0.0, 1.9, 1.3, 2.5, 0.8, 1.8]
# Aphelion in millions of kilometres
aph = [69.8, 108.9, 152.1, 249.2, 816.6, 1514.5, 3003.6, 4545.7]

print("|Name | Inclination | Aphelion | Distance|")
print("|-|-|-|-|")
for n, th, r in zip(names, inc, aph):
    # Perpendicular distance to eciptic
    h = r * sin(radians(th))
    print(f"|{n} | {th} | {r} | {h:.3f}|")

Here's a live version of the script running on the SageMathCell server.

As John Holtz mentions in the comments, the true $h$ value for a planet may be smaller than the value shown in my table. That $h$ value only occurs if the planet's argument of periapsis is ±90°. Fortunately, Neptune's argument of periapsis is currently ~272°, so my $h$ value should be fairly close to the true value.