Answer
${\bf 167}\;\rm ns$
Work Step by Step
According to the given, the two watches are in the same reference frame and one of them is at the origin and the other at point $\rm P=(x,y,z)=(30,40,0) \rm m$ which means in the first quadrant of $x-y$ plane, as shown in the figure below.
Now we need to find the time it takes the light to travel from the original watch to the watch at point $P$.
We know that the speed of light is $c$, so
$$c=\dfrac{r}{t}$$
Hence, the clock should be preset to $t$,
$$t=\dfrac{r}{c}$$
where $r$ is the direct distance between the two clocks which is given by
$$s=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}=\sqrt{x_P^2+y_P^2+z_P^2}$$
$$t=\dfrac{\sqrt{x_P^2+y_P^2+z_P^2}}{c}$$
Plug the known;
$$t=\dfrac{\sqrt{30^2+40^2+0^2}}{(3\times 10^8)}=\color{red}{\bf 167}\;\rm ns$$
