#### Answer

It would take 2.4 seconds to absorb the same amount of energy.

#### Work Step by Step

$U = I~A~t$
$U$ is energy
$I$ is intensity
$A$ is the area
$t$ is the time
Let $P$ be the power of the source. Let $I_1$ be the intensity when the panel is a distance of 1.00 AU from the source. We can find an expression for $I_1$:
$I_1 = \frac{P}{4\pi (1.00~AU)^2}$
$I_1 = 0.07958~P$
We can find an expression for $I_2$:
$I_2 = \frac{P}{4\pi (1.55~AU)^2}$
$I_2 = 0.03312~P$
We can write an expression for the energy collected by the first panel:
$U = I_1~A~t_1$
We can write an expression for the energy collected by the second panel:
$U = I_2~A~t_2$
Since the energy should be equal, we can equate the two expressions to find $t_2$:
$I_2~A~t_2 = I_1~A~t_1$
$t_2 = \frac{I_1~t_1}{I_2}$
$t_2 = \frac{(0.07958~P)(1~s)}{0.03312~P}$
$t_2 = 2.4~s$
It would take 2.4 seconds to absorb the same amount of energy.