Answer
a) $6.40\times10^5 \;\rm kg$
b) $35.7\;\rm year$
Work Step by Step
We know that the final kinetic energy of the spaceship is equal to the energy due to converting mass to energy, so
$$KE=E=mc^2$$
where $m$ is the amount of mass that is converted to energy.
$$\left(\gamma-1\right)m_{1}c^2=mc^2$$
where $m_1$ is the mass of the ship.
$$\left(\gamma-1\right)m_{1} =m $$
$$m=\left(\sqrt{\dfrac{1}{1-\dfrac{v^2}{c^2}}}-1\right)m_{1} $$
Plugging the known;
$$m=\left(\sqrt{\dfrac{1}{1-\dfrac{(0.7c)^2}{c^2}}}-1\right)m_{1}$$
$$m= \left(\sqrt{\dfrac{1}{1- 0.7^2}
}-1\right)\times 160\;000 $$
$$m =\color{red}{\bf 6.40\times10^5}\;\rm kg$$
b) Now we need to find the time the journey takes relative to the onboard astronauts.
We know that the time according to the astronauts is given by
$$\Delta t_{onboard} =\Delta t \sqrt{1-\dfrac{v^2}{c^2}}\tag 1$$
And we know that the time the trip would take for an observer on Earth (outside the system) is given by
$$v=\dfrac{\Delta x}{\Delta t}$$
and hence,
$$\Delta t=\dfrac{\Delta x}{v}$$
Plugging into (1) and then plugging the known;
$$\Delta t_{onboard} =\dfrac{\Delta x}{v} \sqrt{1-\dfrac{v^2}{c^2}} $$
$$\Delta t_{onboard} =\dfrac{35c}{0.7c} \sqrt{1-\dfrac{0.7^2c^2}{c^2}} $$
Noting that the distance of 35 lightyears means the distance traveled by light during one full year which is given by the speed of light times the year.
$$\Delta t_{onboard} = \color{red}{\bf 35.7}\;\rm year $$
