prism ..solid mensuration

To get the height of this weird cylinder, recall the formula:

$$V=BH$$

where $V$ is the volume of the figure, $B$ is the area of the base, and $H$ is the height of the figure.

With that in mind, we plug this given to the formula above:

$$V=BH$$ $$350 \space in^3 = \left(\frac{1}{2}r^2\theta\right)(4r)$$
$$350 \space in^3 = \left(\frac{1}{2}(r^2)\left(25^o\left(\frac{\pi}{180^o}\right)\right)\right)(4r)$$ $$350 \space in^3 = \left(\frac{1}{2}(r^2)\left(\frac{5\pi}{36}\right)\right)(4r)$$ $$350 \space in^3 = \frac{5\pi}{18}r^3 $$ $$\frac{1260}{\pi} = r^3$$ $$r = 7.3 \space inches$$

Note that the area of a sector of circle is $A=\frac{1}{2}r^2\theta$, where $r$ is the radius of the sector and $\theta$ is the central angle of the sector.

Since the problem asks for the height of this weird cylinder, which is four times longer than the radius of the sector of circle, so:
$$H = 4r$$ $$H = 4(7.3 \space in)$$ $$H = 29.2 \space in$$

Therefore, the height of this weird cylinder is $\color{green}{H=29.2 \space in}$.

Alternate solutions are encouraged...