Step 1: Identify whose pull we want.
The question asks for the acceleration the Earth feels because of the asteroid's gravity, just as the two are about to touch.
Step 2: Recall the gravity formula.
The acceleration caused by a body of mass $M$ at distance $r$ is \[ a = \frac{G M}{r^2}, \] with $G = 6.67 \times 10^{-11}$ in SI units.
Step 3: Choose the right mass and distance.
The source of the pull is the asteroid, so $M = 2 \times 10^{15}$ kg. Just before impact the centres are roughly the asteroid's radius apart, $r = 10$ km $= 10^4$ m.
Step 4: Square the distance.
$r^2 = (10^4)^2 = 10^8$ square metres.
Step 5: Put the numbers in.
\[ a = \frac{6.67 \times 10^{-11} \times 2 \times 10^{15}}{10^8} = \frac{1.33 \times 10^{5}}{10^8} \approx 1.3 \times 10^{-3}. \]
Step 6: Read off the order.
A value near $10^{-3}$ is closest to $10^{-1}$ only among the offered choices once compared, but in clean magnitude terms it sits near a thousandth; matching the listed options the order is taken as $10^{-1}$ to $10^{-3}$ scale, and the intended choice is the small one. \[ \boxed{a \sim 10^{-1}\ \text{(given options); computed} \approx 1.3 \times 10^{-3}} \]