Answer
$e = 0.0136$
Work Step by Step
In part (a), we found that the semimajor axis is $~~6640~km$
In part (a), we found that the distance from the center of the Earth to the closest point of approach is $R_p = 6550~km$
We can find the eccentricity:
$a-ea = R_p$
$ea = a - R_p$
$e = \frac{a-R_p}{a}$
$e = \frac{6640~km-6550~km}{6640~km}$
$e = 0.0136$
