Answer
$4.71\times10^{-6} kg \frac{m^2}{s}$.
Work Step by Step
First find the moment of inertia of the second hand, treating it like a rod.
$$I=\frac{1}{3}(0.006kg)(0.15m)^2$$
$$I=4.5\times 10^{-5}kg\cdot m^2$$
Now find the angular momentum.
$$L=I \omega=(4.5\times 10^{-5}kg\cdot m^2)\frac{1rev}{60s}\frac{2 \pi rad}{rev}=4.71\times10^{-6} kg \frac{m^2}{s}$$
