Answer
The comet's greatest distance from the sun is $~~6.4~R_p$
Work Step by Step
In part (a), we found that the semi-major axis is $~~a = 1.9\times 10^{13}~m$
We can find $R_a$, the aphelion distance:
$R_a = ea+a$
$R_a = (0.9932)(1.9\times 10^{13}~m)+(1.9\times 10^{13}~m)$
$R_a = 3.8\times 10^{13}~m$
We can express this distance in terms of $R_p$, the mean orbital radius of Pluto:
$\frac{R_a}{R_p} = \frac{3.8\times 10^{13}~m}{5.90\times 10^{12}~m}$
$R_a = 6.4~R_p$
The comet's greatest distance from the sun is $~~6.4~R_p$
