#### Answer

The power dissipated by each bulb is $~18.75~W$

#### Work Step by Step

We can write an expression for the power:
$P = I^2~R$
We know that the current is $I = \frac{V}{R}$. If two bulbs are connected in series, then the total resistance is doubled, so the current is halved. Then the power dissipated by one bulb is reduced by a factor of $\frac{1}{4}$
The power dissipated by each bulb is $\frac{75~W}{4} = 18.75~W$