The easiest thing to do is to scale from 1 AU

which gives F=50.7 W/m^2 at r=5.2 AU. Note that direct computation gives

F = L_sun / (4r²)

which agrees within our rounding of the solar constant 1370 W/m^2. Thus, to get our 500 W of power, we need an area of around

Area = P / F 500 W / 50 W/m² = 10 m²

or a panel size of 3.3m x 3.3m! This is pretty big (10ft on a side) and is why Galileo carries an onboard nuclear power source.

A flat panel oriented toward the Sun absorbs over the forward half of its surface area only, while it emits thermal radiation over both sides, so the power equilibrium is given by

P_in = F_in L² =
P_out = T⁴ 2L²
=> T = [ F_in/2 ]^1/4 = 145 K.