1. Given data

• Maximum distance from Sun (aphelion): $r_a = 1.6 \times 10^{12}\ \text{m}$
• Minimum distance from Sun (perihelion): $r_p = 8.0 \times 10^{10}\ \text{m}$
• Speed at nearest point: $v_p = 6 \times 10^4\ \text{m/s}$
• Speed at farthest point: $v_a = ?$

2. Use conservation of angular momentum

For motion under gravitational force, angular momentum is conserved. At nearest and farthest points, velocity is perpendicular to radius vector, so:

$m r_p v_p = m r_a v_a$

Cancelling $m$:

$r_p v_p = r_a v_a$

Thus,

$v_a = \frac{r_p v_p}{r_a}$

3. Substitute the values

$v_a = \frac{(8.0 \times 10^{10})(6 \times 10^4)}{1.6 \times 10^{12}}$

First simplify the numerical factor:

$\frac{8.0 \times 6}{1.6} = \frac{48}{1.6} = 30$

Now powers of 10:

$10^{10} \times 10^4 \div 10^{12} = 10^{14-12} = 10^2$

So,

$v_a = 30 \times 10^2 = 3.0 \times 10^3\ \text{m/s}$

4. Match with the options

$3.0 \times 10^3\ \text{m/s}$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They match.