Explanation

(c) It is self-evident that the orbit of the comet is elliptic with sun begin at one of the focus. Now, as for elliptic orbits, according to kepler’s third law,
$T^{2} = \frac{4\pi^{2}a^{3}}{GM} \Rightarrow a = \left( \frac{T^{2}GM}{4\pi^{2}} \right)^{1/3}$
$a = \left\lbrack \frac{\begin{array}{r}
\left( 76 \times 3.14 \times 10^{7} \right) \times 6.67 \times 10^{- 11} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times 2 \times 10^{10}
\end{array}}{4\pi^{2}} \right\rbrack^{1/3}$
But in case of ellipse,
$2a = r_{\min} + r_{\max}$
$\therefore\ r_{\max} = 2a - r_{\min} = 2 \times 2.7 \times 10^{12} - 8.9 \times 10^{10}$
$\cong 5.3 \times 10^{12}m$