Answer
The velocity of particle B as seen by particle A is $-0.99~c$
Work Step by Step
Let $v_{EA}$ be the velocity of the Earth relative to particle A. Then $v_{EA} = -0.90~c$
Note that $v_{BE} = -0.90~c$
We can find $v_{BA}$:
$v_{BA} = \frac{V_{BE}~+~v_{EA}}{1+\frac{(v_{BE})~(v_{EA})}{c^2}}$
$v_{BA} = \frac{(-0.90c)~+~(-0.90c)}{1+\frac{(-0.90c)~(-0.90c)}{c^2}}$
$v_{BA} = \frac{-1.80c}{1.81}$
$v_{BA} = -0.99~c$
The velocity of particle B as seen by particle A is $-0.99~c$.
