Answer
See the detailed answer below.
Work Step by Step
First of all, we need to find the resistance of each lightbulb.
We know that the maximum power dissipated in each bulb that allows it to work without blowing up is given by
$$P=\dfrac{V_{\rm rms}^2}{R}$$
So,
$$R=\dfrac{V_{\rm rms}^2}{P}$$
So,
$$R_1=\dfrac{V_{\rm rms}^2}{P_{(\rm 40 W)}}=\dfrac{(120)^2}{40}=\bf 360\;\rm \Omega$$
$$R_2=\dfrac{V_{\rm rms}^2}{P_{(\rm 60 W)}}=\dfrac{(120)^2}{60}=\bf 240\;\rm \Omega$$
$$R_3=\dfrac{V_{\rm rms}^2}{P_{(\rm 100W)}}=\dfrac{(120)^2}{100}=\bf 144\;\rm \Omega$$
Now to find the actual power dissipated in each bulb, we need to find the $\rm rms$ current that passes through each one of them.
We can see that $R_1$ and $R_2$ are in series, so they have the same current $I_{\rm rms,1}$ while $R_3$ is in parallel to the two in series bulbs, so it has a current of $I_{\rm rms,2}$.
$$I_{\rm rms,1}=\dfrac{\varepsilon_{\rm rms}}{R_1+R_2}=\dfrac{120}{360+240}=\bf \frac{1}{5}\;\rm A$$
$$I_{\rm rms,2}=\dfrac{\varepsilon_{\rm rms}}{ R_3}=\dfrac{120}{144}=\bf \frac{5}{6}\;\rm A$$
The actual power dissipated in each bulb is given by
$$P_1=I_{\rm rms,1}^2R_1=\left(\frac{1}{5}\right)^2(360)=\color{red}{\bf 14.4}\;\rm W$$
$$P_2=I_{\rm rms,1}^2R_2=\left(\frac{1}{5}\right)^2(240)=\color{red}{\bf 9.60}\;\rm W$$
$$P_3=I_{\rm rms,2}^2R_3=\left(\frac{5}{6}\right)^2(144)=\color{red}{\bf 100}\;\rm W$$
