Calculating parallel resistance of a rated bulb
I'll presume your bulbs are ordinary resistors first. The 60 W bulb is R1, the other one R2.

Resistance can be calculated as  

\$ R = \dfrac{V^2}{W} \$  

For R1 and R2 that's 

\$ R1 = \dfrac{(110 V)^2}{60 W} = 202 \Omega \$   

\$ R2 = \dfrac{(110 V)^2}{110 W} = 110 \Omega \$ 

Then Rp || R1 = R2, or

\$\dfrac{Rp \cdot R1}{Rp + R1} = R2 \$

Filling in the values gives

\$\dfrac{Rp \cdot 202 \Omega}{Rp + 202 \Omega} = 110 \Omega \$

Solving for Rp gives us 242 &Omega;.

That would it be if the bulbs were ordinary resistors. They're not they're PTC resistors with a very low resistance at room temperature, and higher when the filament is heated by the current. Operating them at lower voltages by placing two of them in series will decrease the resistance, but you'll have to measure the voltage across each of them ,and the current to know what the resistance at that voltage is. So for the bulbs there's no easy answer.