A) 11.1 cm
We can find the focal length of the lens by using the lens equation:

where
f is the focal length
p = 16.0 cm is the distance of the object from the lens
q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)
Solving the equation for f:

B) Converging
The focal length is:
- Positive for a converging lens
- Negative for a diverging lens
In this case, the focal length is positive, so it is a converging lens.
C) 18.0 mm
The magnification equation states that:

where
is the heigth of the image
is the height of the object


Solving the formula for
, we find

So the image is 18 mm high.
D) Inverted
From the magnification equation we have that:
- When the sign of
is positive, the image is erect
- When the sign of
is negative, the image is inverted
In this case,
is negative, so the image is inverted.