Answer
$5.18\;\rm \mu m/s, 333.4^\circ$
Work Step by Step
We know that the average velocity is given by
$$v_{avg}=\dfrac{\vec R}{t_{tot}}\tag 1$$
whereas $\vec R$ is the resultant displacement and $t_{tot}$ is the time it takes for the entire trip.
First, we need to find $\vec R$.
From the given graph:
$$\vec R=-(\vec{AB}+\vec{BC}+\vec{CD}+\vec{DE})=-\vec{EA}=\vec{AE}$$
$$\vec R=\vec{AE}=-(\vec{AE})_x\hat i-(\vec{AE})_y\hat j$$
$$\vec R= (40\;{\rm \mu m})\hat i+(-20\;{\rm \mu m})\hat j$$
$$\boxed{\vec R= (40\;{\rm \mu m})\hat i-(20\;{\rm \mu m})\hat j}$$
The magnitude of $\vec D$ is given by applying the Pythagorean theorem.
$$|\vec R|=\sqrt{R_x^2+R_y^2}=\sqrt{40^2+(-20)^2}=\color{blue}{\bf 44.7}\;\mu m\tag 2$$
The direction of the average velocity is the direction of the resultant displacement. And the displacement direction is given by
$$\tan\alpha_R=\dfrac{R_y}{R_x}$$
Thus,
$$\alpha_R=\tan^{-1}\left[\dfrac{R_y}{R_x}\right]=\tan^{-1}\left[\dfrac{-20}{40}\right]=\bf -26.6^\circ$$
And since it is in the fourth quadrant;
$$\alpha_R=\color{red}{\bf 333.4^\circ}$$
Now we need to find the total time it takes for the whole trip.
$$t_{tot}=t_{\vec{AB}}+t_{\vec{BC}}+t_{\vec{CD}}+t_{\vec{DE}} $$
And since the velocity of E.coli bacterium is constant, the time it takes for each move is given by
$$t=\dfrac{d}{v}$$
So, we need to find the length of each vector and divide it by its velocity.
$$t_{tot}=\dfrac{\sqrt{50^2+10^2}}{20}+\dfrac{10} {20}+\dfrac{\sqrt{(90-50)^2+(30-20)^2}}{20}+\dfrac{\sqrt{(40-90)^2+(-20-30)^2}}{20}$$
$$t_{tot}=\color{blue}{\bf 8.65}\;\rm \mu s\tag 3$$
Plugging (2) and (3) into (1);
$$v_{avg}=\dfrac{ 44.7}{8.65} =\color{red}{\bf 5.18}\;\rm \mu m/s$$