An "angle of elevation" of $300$ degrees is very non-standard. In the ordinary meaning of the term, the angle of elevation is always less than $90$ degrees. I will assume that $300$ is a typo for $30^\circ$. There is some evidence for that: with an angle of elevation of $30^\circ$, the numbers turn out very nice. We get the good old $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle. and one of the given options is correct. With other angles of elevation, the calculation can be done in a similar way, the numbers are just uglier.
Draw a diagram. Label the top of the pole $T$, the bottom of the pole $B$, and the location of the observer $O$. Drop a perpendicular from the top $T$ of the pole to the line $BT$. Suppose this meets the line segment $BT$ at $S$, the tip of the shadow of the pole.
Look at $\triangle TBS$, and let $\angle TBS=\theta$. Let $p$ be the length of the pole. Then $\cos\theta=\frac{7.5}{p},\quad\text{and therefore}\quad7.5=p\cos\theta. \tag{$1$}$ We are told that $BS=7.5$. Thus $SO=30-7.5=22.5$. Note that $ST=p\sin\theta$. We have $\frac{ST}{SO}=\frac{p\sin\theta}{22.5}=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$ It follows that $p\sin\theta=\frac{22.5}{\sqrt{3}}=7.5\sqrt{3}.\tag{$2$}$ Now $(p\cos\theta)^2+(p\sin\theta)^2=p^2(\cos^2\theta+\sin^2\theta)=p^2.$ We conclude from $(1)$ and $(2)$ that $(7.5)^2+ (7.5\sqrt{3})^2=p^2.$ This simplifies to $(7.5)^2(4)=p^2$, from which it follows that $p=(7.5)(2)=15$.
Remarks: $1.$ It would be more standard, and easier in this case, to observe that $SO=22.5$ and since $\frac{ST}{SO}=\tan(30^\circ)$, we have $ST=(22.5)/\sqrt{3}=7.5\sqrt{3}$. Then by the Pythagorean Theorem, $(BT)^2=(ST)^2+(BT)^2=(7.5\sqrt{3})^2+(7.5)^2$, and we are finished.
$2.$ I did not pay attention to the choices provided in this multiple choice question. However, the answers provided here are quite far from each other. It might be best test-taking strategy to rule out the incorrect answers by making a few sketches. In this case, that is far faster than the method that I used.