$T_t = \dfrac{\text{Length}}{\text{Bandwidth}}$
$\quad =\dfrac{50000\times 8}{10^6}$
$ \quad = \dfrac{40}{100} = 0.4s = 400\; \text{ms}$
$T_p = \dfrac{\text{Distance}}{\text{Velocity}}$
$\quad =\dfrac{10000\times 10^3}{2\times 10^8}$
$ \quad = \dfrac{1}{20} = 0.05s = 50\; \text{ms}$
Hence, answer (D) $p=400,q=50.$