Answer
$4.0\times10^{-7}\;\rm W/m^2$
$0.017 \;\rm V/m $
Work Step by Step
According to the description in the problem and according to our previous knowledge, the intensity for photons is the number of photons striking an area of $1.0\;\rm m^2$ per second.
Thus,
$$I=\dfrac{N}{A}\;E=\dfrac{Nh\nu}{A}$$
where $\nu$ is the frequency.
And hence,
$$I= \dfrac{Nhc}{A\lambda}$$
Plugging the known;
$$I= \dfrac{1\times10^{12}\times 6.626\times10^{-34}\times3\times10^8}{1\times 497\times10^{-9}}$$
$$I=\color{red}{\bf 4.0\times10^{-7}}\;\rm W/m^2$$
Now we need to find the maximum electric field which is given by
$$I=\dfrac{\varepsilon_0 c E^2}{2}$$
Solving for $E$;
$$E=\sqrt{\dfrac{2I}{\varepsilon_0 c}}$$
Plugging the known;
$$E=\sqrt{\dfrac{2\times 4\times10^{-7}}{8.85\times10^{-12}\times3\times10^8}}$$
$$E=\color{red}{\bf 0.017}\;\rm V/m $$
