Answer
$({\bf -1.73\times 10^6}\;\hat i ) \;\rm V/m$
Work Step by Step
From the given graph, we can see that:
- $B= 0.10\;\hat i\;\rm T$
- $\sum F=3.2\times 10^{-13}(-\cos30^\circ\;\hat i+\sin30^\circ\;\hat j)\;\rm N$
- $v=10^7\;\hat z\;\rm m/s$
So the net force exerted on a moving charge in an electromagnetic field is given by
$$\sum F=q(\vec E+\vec v\times \vec B)$$
Solving for $\vec E$;
$$\vec E =\dfrac{\sum F}{q}-\vec v\times \vec B$$
Plug the given;
$$\vec E =\dfrac{3.2\times 10^{-13}(-\cos30^\circ\;\hat i+\sin30^\circ\;\hat j)}{1.6\times 10^{-19}}-(10^7\;\hat z)\times (0.10\;\hat i)$$
$$\vec E =-1.73\times 10^6\;\hat i+ 10^6\;\hat j - 10^6\;\hat j $$
Thus,
$$\vec E =(\color{red}{\bf -1.73\times 10^6}\;\hat i ) \;\rm V/m$$
It is obvious now that the direction of the electric field is to the left, in the negative $x$-axis.
