Answer
The rate at which the earth's radial vector sweeps out area is
$\frac{{dA}}{{dt}} \simeq 2.241 \times {10^9}$ $k{m^2}/s$
The magnitude of the vector ${\bf{J}}$ is
$J \simeq 4.482 \times {10^9}$ $k{m^2}/s$.
Work Step by Step
Since the earth's orbit is circular of radius $R = 150 \times {10^6}$ km, we can parametrize it by
${\bf{r}}\left( t \right) = 150 \times {10^6}\left( {\cos \omega t,\sin \omega t} \right)$
Using the period of the earth $T=365$ days $ = 31.536 \times {10^6}$ s, we find the angular speed:
$\omega = \frac{{2\pi }}{T} = \frac{{2\pi }}{{31.536 \times {{10}^6}}}$ 1/s
So, ${\bf{r}}\left( t \right) = 150 \times {10^6}\left( {\cos \frac{{2\pi }}{{31.536 \times {{10}^6}}}t,\sin \frac{{2\pi }}{{31.536 \times {{10}^6}}}t} \right)$.
It follows that
$r{\left( t \right)^2} = {\bf{r}}\left( t \right)\cdot{\bf{r}}\left( t \right) = {\left( {150 \times {{10}^6}} \right)^2} = 22.5 \times {10^{15}}$ $k{m^2}$
The rate at which the earth's radial vector sweeps out area in units of square kilometers per second is
$\frac{{dA}}{{dt}} = \frac{1}{2}r{\left( t \right)^2}\theta '\left( t \right)$
Since $\theta '\left( t \right) = \omega = \frac{{2\pi }}{{31.536 \times {{10}^6}}}$, so
$\frac{{dA}}{{dt}} = \frac{1}{2} \times 22.5 \times {10^{15}} \times \frac{{2\pi }}{{31.536 \times {{10}^6}}} \simeq 2.241 \times {10^9}$ $k{m^2}/s$
Since $\frac{{dA}}{{dt}} = \frac{1}{2}J$, so $J \simeq 4.482 \times {10^9}$ $k{m^2}/s$.
