Answer
$1000\;\rm V/m$, $127^\circ$ CCW from $+x$axis.
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 $V=150x^2-200y^2$
$$\vec E=-\left[ \dfrac{\partial (150x^2-200y^2)}{\partial x}\;\hat i+\dfrac{\partial (150x^2-200y^2)}{\partial y}\;\hat j\right]$$
$$\vec E=-\left[ 300x\;\hat i-400y\;\hat j\right]$$
$$\boxed{\vec E=- 300x\;\hat i+400y\;\hat j}$$
At $(x,y)=(2\;{\rm m}, 2\;{\rm m})$,
$$\vec E=- 300(2)\;\hat i+400(2)\;\hat j $$
$$\vec E= ( - {\bf600}\;\hat i+{\bf 800}\;\hat j )\;\rm V/m$$
So its magnitude is given by
$$E=\sqrt{E_x^2+E_y^2}=\sqrt{(-600)^2+(800)^2}$$
$$E=\color{red}{\bf 1000}\;\rm V/m$$
Since the $E_x$ is negative while $E_y$ is positive, this is the second quadrant.
Thus, its direction, which is counterclockwise from the positive $x$-axis, is given by
$$\theta=180^\circ-\tan^{-1}\left[ \dfrac{E_y}{|E_x|}\right]$$
$$\theta=180^\circ-\tan^{-1}\left[ \dfrac{800}{600}\right]$$
$$\theta= \color{red}{\bf 127}^{\circ}\tag {CCW from $+x$-axis}$$
