Answer
The star radiates EM energy at a rate of $8.82\times 10^{26}~W$
Work Step by Step
We can convert the distance $r$ to units of meters:
$r = (14\times 10^6~ly) \times \frac{9.46\times 10^{15}~m}{1~ly}$
$r = 1.3244\times 10^{23}~m$
We can find the power of the source:
$I = \frac{P}{A}$
$P = I~A$
$P = I~(4\pi~r^2)$
$P = (4\times 10^{-21}~W/m^2)(4\pi)~(1.3244\times 10^{23}~m)^2$
$P = 8.82\times 10^{26}~W$
The star radiates EM energy at a rate of $8.82\times 10^{26}~W$
