Answer
$412\;\rm \mu s$
Work Step by Step
First of all, we need to sketch the situation to make it easy to solve the problem easily, as seen below.
We know that the speed of sound in air is about 343 m/s at room temperature (20$^\circ $C).
Now we need to find the distance from the source to the right ear $L_R$ and the distance from the source to the left ear $L_L$.
From the geometry of the figure below, it is obvious that,
$$\dfrac{h}{\sin45^\circ}=\dfrac{5}{\sin90^\circ}$$
Hence,
$$h=x=5\sin45^\circ\tag 1$$
where $x=h$ since it is an isosceles triangle.
Now we can say that
$$L_R=\sqrt{h^2+(x-0.1)^2}$$
Plugging from (1);
$$L_R=\sqrt{(5\sin45^\circ)^2+(5\sin45^\circ-0.1)^2}=\bf 4.9298\;\rm m$$
By the same approach,
$$L_L=\sqrt{h^2+(x+0.1)^2}$$
$$L_L=\sqrt{(5\sin45^\circ)^2+(5\sin45^\circ+0.1)^2}=\bf 5.0712\;\rm m$$
We know that the speed is given by $v=d/t$, so
$t_R=\dfrac{L_R}{v_{\rm s}}$, and $t_L=\dfrac{L_L}{v_{\rm s}}$
Hence the difference in arrival times is given by
$$\Delta t=t_L-t_R$$
$$\Delta t=\dfrac{L_L}{v_{\rm s}}-\dfrac{L_R}{v_{\rm s}}$$
Plugging the known;
$$\Delta t=\dfrac{5.0712}{343}-\dfrac{4.9298}{343}=\color{red}{\bf 412}\;\rm\mu s$$
