Answer:
It's indeed safer to suggest a 150 VA transformer. Following table however is the clarification given.
Step-by-step explanation:
For 2 motors with 0.1 A, the Power will be:
P =
![2* 120* 0.1](https://img.qammunity.org/2021/formulas/engineering/college/ksecq35jzvf6x19sds4f0wnksvcmjpldek.png)
=
![24 \ W](https://img.qammunity.org/2021/formulas/engineering/college/rj5or7zfux8l9hjhr7inz40307y3bewi92.png)
For 4 motors with 0.18 A, the Power will be:
P =
![4* 120* 0.18](https://img.qammunity.org/2021/formulas/engineering/college/416c03y8jbvqanf5pjhd8ycj5qzg4auhtu.png)
=
![86.4 \ W](https://img.qammunity.org/2021/formulas/engineering/college/xurv8e2dxnurbegzbmh6thgchpxnc2uscq.png)
As we know, for 6 pilot lamps, the power is "5 W".
So,
The total power will be:
⇒ P =
![24+86.4+5](https://img.qammunity.org/2021/formulas/engineering/college/qm82zc230e9n1d4a028phxr73j8chfxdun.png)
=
![115.4 \ W](https://img.qammunity.org/2021/formulas/engineering/college/rko5vjebr34wv43j6yutpj2zmli7qdz2yi.png)
Now,
Consider the power factor to be "0.95"
VA of transformer is:
=
![PF* Power](https://img.qammunity.org/2021/formulas/engineering/college/i51gzd6d3q82r0i17u5b99eibd9megyez2.png)
=
![115.4* 0.9](https://img.qammunity.org/2021/formulas/engineering/college/2ximdruv3ehgema8sstk6bw9f9p78qlzgw.png)
=
![109.63 \ VA](https://img.qammunity.org/2021/formulas/engineering/college/wz9vnqz4mzggyuh6prtpnt7c41jo1ofq6j.png)