In order to find by how much is the car magnified, first let's find the position of the image using the formula below:
![(1)/(f)=(1)/(d_o)+(1)/(d_i)](https://img.qammunity.org/2023/formulas/physics/college/py056kkxgkcki61yfxyif4uflmt8kbjyix.png)
Where f is the focal length, do is the object position and di is the image position.
After converting all measures to meters, we have:
![\begin{gathered} (1)/(0.0505)=(1)/(0.1923)+(1)/(d_i)\\ \\ 19.80198=5.20021+(1)/(d_i)\\ \\ (1)/(d_i)=14.60177\\ \\ d_i=0.0685\text{ m}=6.85\text{ cm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/4epvg44i9j4ki3a4qpw2996tjzgci1izh3.png)
Now, to find the magnification factor, we use the formula below:
![M=(-d_i)/(d_o)=(-0.0685)/(0.1923)=0.356](https://img.qammunity.org/2023/formulas/physics/college/g0yoafgzg1tdu6rxhcl6fjeru67figmeoy.png)
The car is magnified by a factor of 0.356 (that is, the image is smaller than the object)
Now, for the second part of the question, let's use the first formula again, with f = 0.0316 m and di = 0.1534 m:
![\begin{gathered} (1)/(f)=(1)/(d_o)+(1)/(d_i)\\ \\ (1)/(0.0316)=(1)/(d_o)+(1)/(0.1534)\\ \\ (1)/(d_o)+6.5189=31.6456\\ \\ (1)/(d_o)=25.1267\\ \\ d_o=0.0398\text{ m}=3.98\text{ cm} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/krg30wy64x2pei9a0950tfms7aekyetvez.png)
Therefore the object should be put at 3.98 cm.