Answer
$-0.24N{\hat{y}}$
Work Step by Step
We know that
$E=\frac{F}{q}$
We plug in the known values to obtain:
$E=\frac{0.44}{5.0\times 10^{-6}}=8.8\times 10^4\frac{N}{C}$
Now $F=qE$
We plug in the known values to obtain:
$F=(-2.7\times 10^{-6})(8.8\times 10^4)=-0.24N{\hat{y}}$
