Answer
See the detailed answer below.
Work Step by Step
a) If the mission failed, the asteroid will continue moving toward the Earth at the same speed of 20 km/s (since we ignored Earth's gravitational pull) from a distance of $4\times 10^6$ km.
Thus, the time it takes to hit the Earth is given by
$$v=\dfrac{d}{t}$$
$$t=\dfrac{d}{v}$$
Plugging the known;
$$t=\dfrac{4\times 10^6\;\rm km}{20\;\rm km/s}\times \dfrac{1\rm\;h}{60^2\rm \;s}=\color{red}{\bf 55.56}\;\rm h$$
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b) From a distance of $4\times 10^6$ km, as we see in the figure below, the deflected angle is given by
$$\tan\theta_{minimum}=\dfrac{R_E}{d}$$
$$\theta_{minimum}=\tan^{-1}\left( \dfrac{R_E}{d} \right)$$
Plugging the known;
$$\theta_{minimum}=\tan^{-1}\left( \dfrac{6400}{4\times 10^6} \right)=\color{red}{\bf 0.092^\circ }$$
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c) The mass of the rocket is negligible relative to the mass of the asteroid, so we can assume that the whole thrust of the rocket is perfectly horizontal on the asteroid.
So that the horizontal acceleration component of the asteroid after hitting by the rocket is given by
$$a_x=\dfrac{F_T}{M_{astro}}$$
Plugging the known;
$$a_x=\dfrac{5\times 10^9}{4\times 10^{10}}=\bf 0.125\;\rm m/s^2$$
Now we need to find the horizontal velocity component of the asteroid;
$$v_x=\overbrace{v_{ix}}^{=0}+a_xt=0.125\times 300=\bf 37.5\;\rm m/s$$
And hence, the vertical velocity component remains constant since the rocket affects perfectly horizontal.
Thus,
$$v_y=20\;\rm km/s=\bf 20\times 10^3\;\rm m/s$$
Therefore, the deflection angle due to the thrust of the rocket on the asteroid, as we see in the figure below, is given by
$$\tan\theta =\dfrac{v_x}{v_y}$$
$$\theta =\tan^{-1}\left(\dfrac{v_x}{v_y}\right)=\tan^{-1}\left(\dfrac{37.5}{20\times 10^3}\right)$$
$$\theta=\color{red}{\bf 0.107^\circ}$$
And this deflection angle is greater than the minimum angle required to save the Earth. Hence the Earth would be saved.