Answer
The pulse arrives earliest in tube A and tube B (same arrival time), followed by tube C.
Work Step by Step
We can find an expression for the time of arrival in each tube:
$t_A = \frac{d}{v} = \frac{6L}{\sqrt{\frac{B}{\rho}}} = \frac{6L}{\sqrt{\frac{16B_0}{\rho_0}}} = \frac{3}{2} \times \sqrt{\frac{\rho_0}{B_0}}~L$
$t_B = \frac{d}{v} = \frac{3L}{\sqrt{\frac{B}{\rho}}} = \frac{3L}{\sqrt{\frac{4B_0}{\rho_0}}} = \frac{3}{2} \times \sqrt{\frac{\rho_0}{B_0}}~L$
$t_C = \frac{d}{v} = \frac{2L}{\sqrt{\frac{B}{\rho}}} = \frac{2L}{\sqrt{\frac{B_0}{\rho_0}}} = 2 \times \sqrt{\frac{\rho_0}{B_0}}~L$
We can rank the tubes according to the time of arrival of the pulses at the open right ends of the tubes, earliest first:
The pulse arrives earliest in tube A and tube B (same arrival time), followed by tube C.
