The object/image distances are the wrong way round - as quoted they give a magnification of -2. Correcting that, I agree the values.
"Vignetting" is what happens when the bundle of rays that make it through one optical element don't all make it through the next.
I'm not sure from the description whether we've got object-stop-lens-image or object-lens-stop-image. I'm assuming the former, but extending the argument to the latter isn't hard. In this case, it's convenient to treat the object as positioned at x=0, the stop at x=200 and the lens at x=240. For a point on the object a distance ##h## off axis, it is easy to write down the equations of the rays that pass through the top and bottom of the stop:$$\begin{eqnarray*}
y_t&=&h-\frac{h-10}{200}x\\
y_b&=&h-\frac{h+10}{200}x
\end{eqnarray*}$$Then all you have to do is work out the height of the rays at the position of the lens (x=240):$$\begin{eqnarray*}
y_t&=&12-h/5\\
y_b&=&-12-h/5
\end{eqnarray*}$$and require that they pass through the 25mm diameter lens:$$\begin{array}{rcccl}
-12.5&\leq&12-h/5&\leq&12.5 \\
-12.5&\leq&-12-h/5&\leq&12.5
\end{array}$$
Which comes down to:$$\begin{array}{rcccl}
-0.5&\leq&h/5&\leq&24.5\\
-24.5&\leq&h/5&\leq&0.5
\end{array}$$We require all four of these conditions to be satisfied simultaneously, i.e. ##-0.5\leq h/5\leq 0.5##, or ##-2.5\leq h'\leq 2.5##. So a 5mm object centred in the field of view is not quite affected by vignetting.
In half-angle terms the field of view is simply ##\tan^{-1}(2.5/240)##, the angle subtended by ##h## at the center of the lens.
