A slide projector has a slide 36 mm wide…

Step 1. The magnification. The picture is real, so it is inverted: $m = -33.3$.

Step 2. The distance of the slide. We want $u$, so $u = \dfrac{v}{m} = \dfrac{2.40\ \text{m}}{-33.3} = -0.0720\ \text{m} = -7.20\ \text{cm}$. The slide is $7.20\ \text{cm}$ in front of the lens.

Step 3. The focal length. $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{2.40\ \text{m}} + \dfrac{1}{0.0720\ \text{m}} = 14.31\ \text{m}^{-1}$. So $f = 0.0699\ \text{m} = 6.99\ \text{cm}$.