Answer :
In order to find by how much is the car magnified, first let's find the position of the image using the formula below:
[tex]\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}[/tex]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:
[tex]\begin{gathered} \frac{1}{0.0505}=\frac{1}{0.1923}+\frac{1}{d_i}\\ \\ 19.80198=5.20021+\frac{1}{d_i}\\ \\ \frac{1}{d_i}=14.60177\\ \\ d_i=0.0685\text{ m}=6.85\text{ cm} \end{gathered}[/tex]Now, to find the magnification factor, we use the formula below:
[tex]M=\frac{-d_i}{d_o}=\frac{-0.0685}{0.1923}=0.356[/tex]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:
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\\ \\ \frac{1}{0.0316}=\frac{1}{d_o}+\frac{1}{0.1534}\\ \\ \frac{1}{d_o}+6.5189=31.6456\\ \\ \frac{1}{d_o}=25.1267\\ \\ d_o=0.0398\text{ m}=3.98\text{ cm} \end{gathered}[/tex]Therefore the object should be put at 3.98 cm.