 Solution

Use the Earth's inertial frame, with the Earth at $x=0$ and the star at $x=D=4.2$ light years, neglecting their relative motion. Take the stated relative velocity of $B$ to be forward along the same direction as $A$. The relativistic velocity addition law givesIn units of light years and years, $c=1$ and the spacecraft worldlines are $x_A=t/4$ and $x_B=2t/3$. Their passages at the star occur atThe return radio wave from $B$ follows the leftward light worldlineIntersecting this with $A$'s worldline gives $t/4=10.5-t$, henceThis is the Earth-frame coordinate time of reception on $A$, as requested; it is not the reading of $A$'s own clock. If an observer on the Earth watched that reception through a returning light signal, they would see it later.

Each transmission from the star takes another $D/c=4.2$ years to reach the Earth. ThereforeAll times are measured from the common passage by the Earth; the spacecraft continue at their stated velocities during the calculation.