Answer
$A = 9.7\times 10^5~m^2$
Work Step by Step
Let $r$ be the distance of the spaceship from the sun.
We can equate the two equations for force to find the required area $A$:
$\frac{2IA}{c} = \frac{G~M_s~M}{r^2}$
$\frac{2PA}{4\pi~r^2~c} = \frac{G~M_s~M}{r^2}$
$A = \frac{2\pi~c~G~M_s~M}{P}$
$A = \frac{(2\pi)~(3.0\times 10^8~m/s)~(6.67\times 10^{-11}~N~m^2/kg^2)~(2.0\times 10^{30}~kg)~(1500~kg)}{3.90\times 10^{26}~W}$
$A = 9.7\times 10^5~m^2$
