Answer
The speed of electron B in a frame of reference in which electron A is at rest is $\frac{5}{13}~c$
Work Step by Step
Let west be the positive direction.
Let $v_{BL}$ be the velocity of electron B relative to the lab. Then $v_{BL} = \frac{4}{5}~c$
Let $v_{AL}$ be the velocity of electron A relative to the lab. Then $v_{AL} = \frac{3}{5}~c$. Then $v_{LA} = -\frac{3}{5}~c$
We can find $v_{BA}$:
$v_{BA} = \frac{V_{BL}~+~v_{LA}}{1+\frac{(v_{BL})~(v_{LA})}{c^2}}$
$v_{BA} = \frac{(\frac{4}{5}~c)~+~(-\frac{3}{5}~c)}{1+\frac{(\frac{4}{5}~c)~(-\frac{3}{5}~c)}{c^2}}$
$v_{BA} = \frac{\frac{1}{5}~c}{\frac{13}{25}}$
$v_{BA} = \frac{5}{13}~c$
The velocity of electron B relative to electron A is $\frac{5}{13}~c$
Therefore, the speed of electron B in a frame of reference in which electron A is at rest is $\frac{5}{13}~c$
(to the west)
