Calculating the length of one year for a sun-orbiting planet?!?

6.95 x 10^8 km is the same as 4.6458 AU
4.6458^1.5 = 10.014 years

Keplers third rule

Consider a stylized solar system with negligible mass planets rotating in a stable circular orbit.
Then the following constant serves any of the planets.

k= t ² / r ³

t = time period of 360 ° orbit
r = radius of orbit

Notes :
Use same units for all planets ( seconds/seconds , days/days, years/years – metres/metres, AU/AU etc. ).
This equation can be used in real solar systems in an approximate fashion.

So, if they both orbit our sun, then use the earth as the base :

r = earths orbital radius = 1.5 E+8 km
t = earths orbit time = 31,556,952 seconds
R = planet x orbital radius = 6.95 E+8 km
T = planet x orbit time = ?

From keplers third law, we get :
t ² / r ³ = T ² / R ³
Transpose for T :
T = square root ( ( R ³ * t ² ) / r ³ )
You can calculate the answer

p^2 is proportional to a^3

First, use the information for the earth

a = 1.5 * 10^8 km
p = 365.25 * 24 * 60 * 60

p1^2 / a1^3 = p2^2 / a2^3
(6.95 * 10^8)^3 * (365.25 * 24 * 60 * 60)^2 / (1.5 * 10^8)^3 = p^2
((6.95 * 10^8) / (1.5 * 10^8))^3 * (365.25 * 24 * 60 * 60)^2 = p^2
p = 365.25 * 24 * 60 * 60 * (6.95 / 1.5)^(3/2)
p = 314734452.28074844491631998183197

314734452 seconds

You are asking for the (perimeter) of an Ellipse of eccentricity e. It is very involved subject & is difficult to find out, via a difficult integration process. "Peirce Tables" gives it as a two page tabulation, for various segments of the Ellipse arc for various e.
What I do is this:
Since circumference, C for a circle is 2πr, I hazarded a guess & wrote
C = 2π√[ab] = 2πa√√[1 -e²] = Cₒ[1 -e²]¹/⁴.
For Earth orbit the factor
[1 -e²]¹/⁴= 0.999999302
or the error if I assume it to be circular orbit of radius a, that is Cₒ is about 7 X 10‾⁷.
Oops, you asked for an arc length for 1 year, of any planet that may have its orbital period not necessarily 1 year. The fresh problem that arises is various parts of the orbit have various lengths of orbit. You need to imaginatively use Tables like the Pierce Tables, I mentioned above. This was the toughest problem that only Kepler tackled.