A plano-convex lens of glass with refractive index 1.5 has a curved face of radius 30 cm…

Step 1. The focal length. $\dfrac{1}{F} = \dfrac{2(n - 1)}{R} = \dfrac{2 \times 0.5}{30\ \text{cm}} = \dfrac{1}{30\ \text{cm}}$, so $F = 30.0\ \text{cm}$.

Step 2. The image distance. We want $v$, so $\dfrac{1}{v} = \dfrac{1}{F} - \dfrac{1}{u} = \dfrac{1}{30\ \text{cm}} - \dfrac{1}{45\ \text{cm}} = \dfrac{1}{90\ \text{cm}}$. So $v = 90.0\ \text{cm}$, in front of the lens.

Step 3. The magnification. $m = -\dfrac{v}{u} = -\dfrac{90.0\ \text{cm}}{45\ \text{cm}} = -2.00$. The image is real, inverted and twice as tall.