Answer
$2.3mN$
Work Step by Step
We can find the required average force as follows:
$F_{avg}=P_{avg}A$
$\implies F_{avg}=\frac{I_{avg}A}{c}$ because $P_{avg}=\frac{I_{avg}}{c}$
We plug in the known values to obtain:
$F_{avg}=\frac{675\times 10^3}{3\times 10^8}$
$F_{avg}=2.3mN$
