We are given the following probability density function
$$p(x) = 4 x^{2} \frac{e^{\frac{-2x}{a_{0}}}}{a_{0}^{3}},$$
where $x \geq 0$ and $a_{0} = 5.59 \cdot 10^{-11}$.
The question asks to show that the pdf $p$ is a gamma pdf.
And the gamma pdf is given by
$$f(x) = \frac{b}{\Gamma(a) \langle X \rangle}(\frac{bx}{\langle X \rangle})^{a - 1}e^{\frac{-bx}{\langle X \rangle}} \quad x \geq 0.$$
The notation $\langle X \rangle$ means the mean of the r.v. $X$. I calculated this and it yields $\frac{3}{2 a_{0}}$. However, I am not sure how to show that $p$ is a gamma pdf.
The gamma function is defined by
$$\Gamma(a) = \int_{0}^{\infty} y^{a - 1}e^{-y}dy.$$
There is also a hint given: $\Gamma(n + 1) = n!$
Let $k > 0$ be the shape and $\theta > 0$ the scale of a Gamma distribution $\text{Gamma}(k,\theta)$, whose probability density function is given by $$p(x)=\frac{1}{\Gamma(k)\theta^k}x^{k-1}\exp\left\{-\frac{x}{\theta}\right\}$$ If we take $k=3$ and $\theta=\tfrac{a_0}{2}$, then \begin{equation*} \begin{split} p(x) &=\frac{1}{\Gamma(3)\cdot\left(\tfrac{a_0}{2}\right)^3}x^{3-1}\exp\left\{-\frac{x}{\tfrac{a_0}{2}}\right\}=\frac{8}{\Gamma(2+1)a_0^3}x^{2}\exp\left\{-\frac{2x}{a_0}\right\} \\ &=\frac{8}{2!a_0^3}x^{2}\exp\left\{-\frac{2x}{a_0}\right\}=\frac{4}{a_0^3}x^{2}\exp\left\{-\frac{2x}{a_0}\right\} \end{split} \end{equation*} which is the expression you have