In this question we have used the concept of pipelining.
In second and Third case, First packet will take $3\times T_t$ time and all subsequent
packets will be delivered in one $T_t$ time.
$T_1=3\times T_t=3\times \dfrac{(1000+ 100)}{B}$
$T_t=\dfrac{\text{(data + header)}}{\text{Bandwidth}}$
data $= 1000\text{ Bytes};$ header $=100\text{ Bytes}$
$T_1 =\dfrac{3300}{B}\text{ seconds}$
$T_2 = 3\times T^{'}_{t} +9\times T^{'}_{t}=12\times T^{'}_{t} $
$T^{'}_{t}=\dfrac{\text{(data + header)}}{\text{Bandwidth}}$
$T_2=\dfrac{12\times (100 + 100)}{B}=\dfrac{2400}{B}\text{ seconds}$
$T_3 =3\times T^{"}_{t}+ 19\times T^{"}_{t}=22\times T^{"}_{t}$
$T^{"}_{t}=\dfrac{(50 + 100)}{B}$
$T_3 =\dfrac{22\times 150}{B}=\dfrac{3300}{B}$
So $T_1 = T_3$ and $T_3 > T_2;$
option D