Answer
$(-2.0\times 10^3N/C)\hat x+(3.2\times 10^3N/C)\hat y$
Work Step by Step
We know that
$E^{\rightarrow}=(v^{\rightarrow}\times B^{\rightarrow})$
We plug in the known values to obtain:
$E^{\rightarrow}=[(4.4\times 10^3m/s)\hat x+(2.7\times 10^3m/s)\hat y]\times (0.73)T \hat z$
This simplifies to:
$E^{\rightarrow}=(-2.0\times 10^3N/C)\hat x+(3.2\times 10^3N/C)\hat y$
