Answer
a) ${\bf 2.19\times 10^{11}}\;\rm V/m$
b) $\bf 0.427$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the laser beam is an electromagnetic wave which has an intensity of $I$ that is given by
$$I=\dfrac{P}{A_{\rm lens}}=\dfrac{c\epsilon_0 E_0^2}{2}$$
Assuming that the energy is uniformly distributed through a cross-sectional area of the laser beam.
Solving for $E_0$;
$$ E_0=\sqrt{\dfrac{2P}{\pi c\epsilon_0 r^2_{\rm lens}} }$$
Plug the known;
$$ E_0=\sqrt{\dfrac{2(200\times 10^6)}{\pi (3\times 10^8)(8.85\times 10^{-12}) (1\times 10^{-6})^2 } }$$
$$E_0=\color{red}{\bf 2.19\times 10^{11}}\;\rm V/m$$
$$\color{blue}{\bf [b]}$$
The ratio needed is
$$\dfrac{E_0}{E_{atom}}=\dfrac{ 2.19\times 10^{11}}{E_{atom}}$$
Recalling that the electric field between a proton and an electron is given by
$$E=\dfrac{e}{4\pi \epsilon_0 r^2}$$
So,
$$\dfrac{E_0}{E_{atom}}=\dfrac{4\pi \epsilon_0 r^2( 2.19\times 10^{11})}{e}$$
Plug the known;
$$\dfrac{E_0}{E_{atom}}=\dfrac{ (0.053\times 10^{-9})^2( 2.19\times 10^{11})}{(9\times 10^9)(1.6\times 10^{-19})}$$
$$\dfrac{E_0}{E_{atom}}=\color{red}{\bf 0.427}$$
