Answer
We have $||{\bf{J}}|| = J = r{\left( t \right)^2}\theta '\left( t \right)$.
We show that $\frac{{dA}}{{dt}} = \frac{1}{2}J$.
So, the rate at which the radial vector sweeps out area is half of $J$.
Work Step by Step
The area $A$ swept out by the planet's radial vector is
$A = \frac{1}{2}\mathop \smallint \limits_0^\theta {r^2}{\rm{d}}\theta $
It is shown on page 749 the rate is
$\frac{{dA}}{{dt}} = \frac{1}{2}r{\left( t \right)^2}\theta '\left( t \right)$
By Eq. (3) we have $||{\bf{J}}|| = J = r{\left( t \right)^2}\theta '\left( t \right)$. Thus, $\frac{{dA}}{{dt}} = \frac{1}{2}J$.
So, the rate at which the radial vector sweeps out area is half of $J$.
