1. Understand the configuration

Two identical stars each have mass $M=3\times 10^{31}\ \text{kg}$ and the distance between them is $2a=2\times 10^{11}\ \text{m}$ So each star is at distance $a=10^{11}\ \text{m}$ from the common centre of mass $O$.

The meteorite passes through $O$ along a direction perpendicular to the plane of rotation.

2. Gravitational potential at the centre $O$

Potential due to one star at distance $a$ is $V_1=-\frac{GM}{a}$ Since there are two identical stars, total potential at $O$ is $V=-\frac{GM}{a}-\frac{GM}{a}=-\frac{2GM}{a}$

Substitute values: $V=-\frac{2(6.67\times 10^{-11})(3\times 10^{31})}{10^{11}}$

$V=-2\times 6.67\times 3\times 10^{-11+31-11}$

$V=-40.02\times 10^9=-4.002\times 10^{10}\ \text{J/kg}$

3. Condition for escape

For minimum escape speed, total mechanical energy at $O$ must be zero: $\frac{1}{2}v_e^2 + V = 0$

So, $\frac{1}{2}v_e^2 = -V = \frac{2GM}{a}$

Hence, $v_e^2=\frac{4GM}{a}$

Therefore, $v_e=\sqrt{\frac{4GM}{a}}=2\sqrt{\frac{GM}{a}}$

4. Calculate the value

$v_e=\sqrt{\frac{4(6.67\times 10^{-11})(3\times 10^{31})}{10^{11}}}$

$v_e=\sqrt{80.04\times 10^9}$

$v_e=\sqrt{8.004\times 10^{10}}$

$v_e\approx 2.83\times 10^5\ \text{m/s}$

5. Match with options

$v_e\approx 2.8\times 10^5\ \text{m/s}$

So the correct option is:

D: $2.8\times 10^5\ \text{m/s}$

6. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.