Answer
$\vec{E}_{net} = 40$ N/C
$\theta = 27^{\circ}$
Work Step by Step
We first need to find an equation for the electric field of the region of space. We can do this by taking the derivative of $V(x,y)$ two different times-once with respect to $x$ and once with respect to $y$-to find the $x$ and $y$ components of the electric field. Know that $\displaystyle \vec{E} = -\frac{dV}{d\vec{s}}$, where $\vec{s}$ is the position vector.
$\displaystyle \vec{E}_x(x,y) = -\frac{d}{dx}\left[\frac{200}{\sqrt{x^2+y^2}}\right] = -\frac{d}{dx}\left[200(x^2+y^2)^{-1/2}\right]$
$\displaystyle \vec{E}_x(x,y) =-\left[-100(x^2+y^2)^{-3/2}(2x+0)\right] = \frac{200x}{(x^2+y^2)^{3/2}}$
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$\displaystyle \vec{E}_y(x,y) = -\frac{d}{dy}\left[\frac{200}{\sqrt{x^2+y^2}}\right] = -\frac{d}{dy}\left[200(x^2+y^2)^{-1/2}\right]$
$\displaystyle \vec{E}_y(x,y) =-\left[-100(x^2+y^2)^{-3/2}(0+2y)\right] = \frac{200y}{(x^2+y^2)^{3/2}}$
We can now find the values for the electric field components at the position ($x$, $y$) = ($2$m, $1$m)
$\displaystyle \vec{E}_x(2,1) = \frac{200(2)}{(2^2+1^2)^{3/2}} \approx 35.8$ N/C
$\displaystyle \vec{E}_y(2,1) = \frac{200(1)}{(2^2+1^2)^{3/2}} \approx 17.9$ N/C
Now we can take the magnitude of these components to find $\vec{E}_{net}$
$\vec{E}_{net} = \displaystyle \sqrt{\left(\frac{400}{5^{3/2}}\right)^2+\left(\frac{200}{5^{3/2}}\right)^2} = 40$ N/C
We can find the angle counterclockwise to the $x$-axis by:
$\theta = \displaystyle \tan^{-1}\left(\frac{\vec{E}_y(2,1)}{{\vec{E}_x(2,1)}}\right) = \tan^{-1}\left(\frac{17.9}{35.8}\right) = \tan^{-1}\left(\frac{1}{2}\right) \approx 27^{\circ}$
