Answer
$E=0.0833 \mathrm{~V} / \mathrm{m} .$
Work Step by Step
The magnitude of the electric field at point $P$ is
$$
E=\frac{V}{l}=\frac{i R}{l}=(25.0 \mathrm{~A})\left(\frac{1.00 \Omega}{300 \mathrm{~m}}\right)\\=0.0833 \mathrm{~V} / \mathrm{m} .
$$
The direction of $\vec{E}$ at point $P$ is in the $+x$ direction, same as the current.
