Answer
a) $4.244\times10^3\;\rm W/m^2$, $3.3\dot3\times10^{-9 }\;\rm N$
b) $8.488\times10^6\;\rm m/s^2$
Work Step by Step
a) We know that the pressure force is given by
$$P=\dfrac{F}{A}$$
Thus,
$$F=PA\tag 1$$
In our case, the pressure is given by
$$P=\dfrac{I_{avg}}{c}$$
where $I_{avg}=\dfrac{P_{avg}}{A}$
Hence,
$$P=\dfrac{P_{avg}}{Ac}\tag 2$$
Plugging into (1);
$$F=\dfrac{P_{avg}}{Ac}A=\dfrac{P_{avg}}{c}$$
Plugging the known;
$$F=\dfrac{1}{3\times10^8}=\color{red}{\bf 3.3\dot3\times10^{-9}}\;\rm N$$
The radiation pressure is given by equation (2).
So we just need to plug the known into it.
$$P=\dfrac{P_{avg}}{Ac}=\dfrac{P_{avg}}{\pi r^2c}$$
$$P=\dfrac{1}{3\times10^{8} \cdot \pi\cdot (5\times10^{-7})^2} =\color{red}{\bf4.244\times10^3}\;\rm W/m^2$$
Notice that $P$ is the pressure while $P_{avg}$ is the power, so do not be confused.
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b)
According to Newton's second law, acceleration is given by
$$a=\dfrac{F}{m}$$
and we know that the mass of this water cylinder is given by the density law; $m=\rho V=\rho Ah$.
And in this case, the cross-sectional area is $A=\pi r^2$ and $h=r$.
Therefore,
$$a=\dfrac{F}{\rho \pi r^3}$$
Plugging the known;
$$a=\dfrac{3.3\dot3\times10^{-9} }{1000\cdot \pi\cdot (5\times10^{-7})^3}$$
$$a=\color{red}{\bf8.488\times10^6}\;\rm m/s^2$$