Explanation:
The expression is :

A =[LT], B=[L²T⁻¹], C=[LT²]
Using dimensional of A, B and C in above formula. So,
![A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}](https://tex.z-dn.net/?f=A%3DB%5EnC%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3D%5BL%5E2T%5E%7B-1%7D%5D%5En%5BLT%5E2%7D%5D%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%7DT%5E%7B-n%7DL%5EmT%5E%7B2m%7D%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%2Bm%7DT%5E%7B2m-n%7D)
Comparing the powers both sides,
2n+m=1 ...(1)
2m-n=1 ...(2)
Now, solving equation (1) and (2) we get :

Hence, the correct option is (E).
To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.
Kepler's third law tells us that

Where
T= Period
G= Gravitational constant
M = Mass of the sun
a= The semimajor axis of the comet's orbit
The period in years would be given by

PART A) Replacing the values to find a, we have




Therefore the semimajor axis is 
PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by


