= Solution
For a <Lorentz transformation> with $S'$ moving at $+v$ along $x$ relative to $S$,
$$
x'=\gamma(x-vt),\qquad y'=y,\qquad
t'=\gamma\left(t-\frac{vx}{c^2}\right),\qquad
\gamma=(1-v^2/c^2)^{-1/2}.
$$
Differentiate along the object's trajectory and divide the spatial differentials by $dt'$. Using $u_x=dx/dt$ and $u_y=dy/dt$ derives the <relativistic velocity-addition formula>:
$$
\boxed{u'_x=\frac{u_x-v}{1-u_xv/c^2},\qquad
u'_y=\frac{u_y}{\gamma(1-u_xv/c^2)}.}
$$
In the rest frame of ship A, take $v=c/2$ and the Earth-frame <velocity> of B as $(0,-c/2)$. Here $\gamma=2/\sqrt3$ and the common denominator is one, so
$$
u'_{Bx}=-\frac c2,\qquad u'_{By}=-\frac{\sqrt3}{4}c.
$$
The required relative <speed> is the magnitude of this two-component <velocity>, not either component separately:
$$
\boxed{|\mathbf u'_B|=c\sqrt{\frac14+\frac3{16}}=\frac{\sqrt7}{4}c\simeq0.6614c.}
$$
The inverse <Lorentz factor> check is $\gamma_{AB}=\gamma_A\gamma_B(1-\mathbf u_A\cdot\mathbf u_B/c^2)=4/3$, which gives the same relative <speed>.
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