The given situation is shown below



$a_{\max }=1.4 \times 10^{12} \mathrm{~m}$
$a_{\min }=7 \times 10^{10} \mathrm{~m}$
The velocity of the comet is maximum, when it is nearest to the sun and minimum when it is farthest from the sun.
$\therefore \quad v_{\max }=6 \times 10^{14} \mathrm{~m} / \mathrm{s}$
Applying the law of conservation of angular momentum at points $A$ and $B$, we get
$$
\begin{aligned}
&m v_{\min } a_{\max }=m v_{\max } a_{\min } \quad(\quad L=m v r) \\
&\Rightarrow \quad v_{\min }=\frac{v_{\max } \times a_{\min }}{a_{\max }} \\
&=\frac{6 \times 10^{14} \times 7 \times 10^{10}}{1.4 \times 10^{12}} \\
&=3 \times 10^{13} \mathrm{~ms}^{-1}
\end{aligned}
$$
No option is correct.