Answer
${\bf 9.76}\;\rm h$
Work Step by Step
In this problem, we will use two of Newton's laws, the second and third, and two of the kinematic formulas.
In space, we can assume that the only force exerted on the astronaut in his situation is the reaction force from the radiation force from the laser portable.
$$F_{net}=|-F_{\rm rad}|=ma$$
Hence,
$$a=\dfrac{F_{\rm rad}}{m}\tag 1$$
We know that the radiation pressure on an object that absorbs all the radiation is given by
$$P=\dfrac{I}{c}\tag 2$$
where $P$ here is for pressure.
The force exerted by this radiation is then given by
$$F=PA$$
where $P$ is the pressure and $A$ is the cross-sectional area of the object.
Plug from (2),
$$F=\dfrac{I}{c}A$$
where $I={\rm Power}/A$,
$$F=\dfrac{ { P\rm ower}}{cA}A$$
$$F_{\rm rad}=\dfrac{ {P\rm ower}}{c } $$
Plug into (1),
$$a=\dfrac{{P\rm ower}}{mc}=\dfrac{(1000)}{(80)(3\times 10^8)}$$
$$a=\bf\frac{1}{24,000,000}\;\rm m/s^2 \tag 3$$
According to the kinematic formula, the final velocity after 1 hour of acceleration is given by
$$v_f=v_i+at$$
where the astronaut starts from rest and accelerates for 1 hour since the portable laser can operate for 1 h.
$$v_f=at=a(1\times 3600)$$
Plug from (3)
$$v_f=\dfrac{3600\;({P\rm ower})}{mc}=\dfrac{3600\;(1000)}{(80)(3\times 10^8)}$$
$$v_f= \bf \frac{3}{20,000}\;\rm m/s\tag 4$$
This is the speed at which the astronaut will travel through the rest of the trip.
So we need to find the distance traveled during this acceleration interval $d_1$.
$$v_f^2=v_i^2+2ad_1$$
$$d_1=\dfrac{v_f^2}{2a}=\dfrac{}{}$$
Plug from (3) and (4),
$$d_1=\dfrac{(3/20,000)^2}{2(1/24,000,000)}=\bf 0.27\;\rm m$$
This distance takes a time of 1 h, so we need to find the time the rest of the 5 meters would take at this final speed,
$$v_f=\dfrac{5-d_1}{t_1}$$
So,
$$t_1=\dfrac{5-d_1}{v_f}=\dfrac{5-0.27}{(3/20,000)}=31,533\;\rm s=\bf 8.76\;\rm h$$
Hence the whole trip would take,
$$t_{tot}=1+8.76=\color{red}{\bf 9.76}\;\rm h$$
which is less than the 10-h supply of oxygen, so he can do it.
He will survive.