Let the speed of the bus during the first hour be $t\,kmph$
So, speed during second hour=$2t\,kmph$ and the speed during third hour works out to be $\frac{2}{3} \times (t+2t)=2t\,kmph$
So, total distance travelled by the bus in 3 hours=$t+2t+2t=5t\,km$
Had the bus travelled for 3 hours at the speed of the first hour it would have travelled 120km less
$5t-3t=120\,km$
$t=60$
So, speed during first,second and third hours are $60kmph,120kmph,120kmph$ respectively.
Average speed=$\frac{60+120+120}{3}=100kmph(A)$