Answer
See the detailed answer below.
Work Step by Step
We know that the 2p radial wave function is given by:
$$
R_{2p}(r) = A_{2p} \left( \frac{r}{2a_B} \right) e^{-r/2a_B}\tag 1
$$
And we know that the normalization condition for the three-dimensional hydrogen atom is:
$$
\int_0^\infty P_r(r) dr = 4\pi \int_0^\infty r^2 |R_{2p}(r)|^2 dr = 1
$$
Substituting $ R_{2p}(r) $ from (1);
$$
P_r(r) = \color{red}{\bf\not} 4\pi A_{2p}^2 \int_0^\infty \left( \frac{r^2}{\color{red}{\bf\not} 4a_B^2} \right) e^{-r/a_B} r^2 dr=1
$$
$$
\frac{ \pi A_{2p}^2}{ a_B^2} \int_0^\infty r^4 e^{-r/a_B}=1\tag 2
$$
Use the given hint; $ \int_0^\infty x^n e^{-\alpha x} \, dx $:
$$
\int_0^\infty x^n e^{-\alpha x} dx = \frac{n!}{\alpha^{n+1}}
$$
where $ n = 4 $, $ \alpha = \dfrac{1}{a_B} $
So the integral becomes:
$$
\int_0^\infty r^4 e^{-r/a_B} dr = \frac{4!}{\left( \dfrac{1}{a_B}\right)^5} = 24\; a_B^5
$$
Plug into (2);
$$
\frac{ \pi A_{2p}^2}{ a_B^2} \cdot 24\; a_B^5=1
$$
$$
24\pi A_{2p}^2 a_B^3=1
$$
Solving for $ A_{2p} $:
$$
\boxed{A_{2p} = \left( 24\pi a_B^3 \right)^{-1/2}}
$$
