Answer
$40\;\rm V$, $\approx 27^\circ$
Work Step by Step
We know that the electric field is given by
$$\vec E=-\left[ \dfrac{\partial V}{\partial x}\;\hat i+\dfrac{\partial V}{\partial y}\;\hat j\right]$$
where $V=\dfrac{200}{\sqrt{x^2+y^2}}=200(x^2+y^2)^{-\frac{1}{2}}$
$$\vec E=-\left[ \dfrac{\partial [200(x^2+y^2)^{-\frac{1}{2}})]}{\partial x}\;\hat i+\dfrac{\partial [200(x^2+y^2)^{-\frac{1}{2}}]}{\partial y}\;\hat j\right]$$
$$\vec E=-\left[ 200 (-\frac{1}{2}) (2x)(x^2+y^2)^{-3/2}\;\hat i+200 (-\frac{1}{2}) (2y)(2^2+1^2)^{-3/2}\;\hat j\right]$$
$$\boxed{\vec E= 200x(x^2+y^2)^{-3/2}\;\hat i+200y(x^2+y^2)^{-3/2}\;\hat j}$$
At $(x,y)=(2\;{\rm m}, 1\;{\rm m})$,
$$\vec E= 200(2)(2^2+1^2)^{-3/2}\;\hat i+200(1)(x^2+y^2)^{-3/2}\;\hat j$$
$$\vec E= ( {\bf 35.777}\;\hat i+{\bf 17.888}\;\hat j )\;\rm V/m$$
So its magnitude is given by
$$E=\sqrt{E_x^2+E_y^2}=\sqrt{(35.777)^2+(17.888)^2}$$
$$E=\color{red}{\bf 40}\;\rm V/m$$
Since the $E_x$ is positive and $E_y$ is also positive, this is the first quadrant.
Thus, its direction, which is counterclockwise from the positive $x$-axis, is given by
$$\theta= \tan^{-1}\left[ \dfrac{|E_y|}{|E_x|}\right]$$
$$\theta= \tan^{-1}\left[ \dfrac{|17.888|}{|35.777|}\right]$$
$$\theta= \color{red}{\bf 26.6}^{\circ}\tag {CCW from $+x$-axis}$$
