A small, totally reflecting plane mirror is positioned horizontally...

Let $P$ be the power of the source. The power of the light that passes through the lens and is incident on the mirror is $P' = 0.30 P$.

Since the mirror is totally reflecting, the force exerted by the light beam on the mirror is given by the rate of change of momentum of the photons:

$F = \dfrac{2P'}{c} = \dfrac{2(0.30 P)}{c} = \dfrac{0.60 P}{c}$

For the light to support the weight of the mirror, this force must be equal to the gravitational force acting on the mirror:

$F = mg$

Substituting the given values ($m = 20 \text{ g} = 0.02 \text{ kg}$, $g = 10 \text{ m s}^{-2}$, and $c = 3 \times 10^8 \text{ m s}^{-1}$):

$\dfrac{0.60 P}{3 \times 10^8} = 0.02 \times 10$

$\dfrac{0.60 P}{3 \times 10^8} = 0.2$

$0.60 P = 0.2 \times 3 \times 10^8$

$0.60 P = 0.6 \times 10^8$

$P = 1.0 \times 10^8 \text{ W}$