Step 1: Use momentum conservation of the light itself. Only the half of the emitted power that enters the cup interacts, carrying power $W/2$. Let the axis point toward the pole of the hemisphere.
Step 2: The net axial momentum rate of the incident (interacting) light is $\dfrac{1}{c}\displaystyle\int \frac{W}{4\pi}\cos\theta\, d\Omega$ over the hemisphere, which equals $\dfrac{W}{4\pi c}(2\pi)\left(\tfrac{1}{2}\right) = \dfrac{W}{4c}$ along $+z$.
Step 3: Perfect reflection reverses every radial ray, so the reflected light carries the same axial momentum rate but along $-z$, that is $-\dfrac{W}{4c}$.
Step 4: The rate of change of the light momentum is the final minus the initial value:
\[\Delta \dot p = -\frac{W}{4c} - \frac{W}{4c} = -\frac{W}{2c}\]
directed along $-z$.
Step 5: By Newton's third law the force on the hemisphere is equal and opposite, of magnitude
\[\boxed{F = \dfrac{W}{2c}}\]