Radial Distribution Function $(4πr^2R^2(r))$ gives the probability of the electron being present at a distance $r$ from the nucleus.
Answer: The given statement is correct.
My Query: According to me, isn't the Radial Probability Distribution Function ($RPDF$) used for finding the probability of an electron being present at a distance $r$ from the nucleus within a region of thickness $dr$?
Therefore, shouldn't the above statement be wrong?
Conceptually you are right as the commenters have mentioned, but since we are on a thread about nitpicking, we might as well go the extra distance.
Technically, $r^2R^2$ itself is not a probability but a probability density. In order to get the actual probability, you need to integrate it over a region. The probability of finding an electron between $r = r_1$ and $r = r_2$ would therefore be
$$P = \int_{r_1}^{r_2}r^2R^2 \, \mathrm dr \tag{1}$$
For an infinitesimal region, it suffices to multiply by $\mathrm dr$, such that the infinitesimal probability of finding an electron in the region $(r, r + \mathrm dr)$ is
$$\mathrm dP = r^2R^2 \,\mathrm dr \tag{2}$$
Notice how if you integrate $(2)$ over a finite region you get $(1)$.
Also, some books give this extra factor of $4\pi$, which I never really understood. You have two choices: either the probability density function is $r^2R^2$ (which works for all orbitals), or $4\pi r^2\psi^2$ (which works only for spherically symmetric orbitals, i.e. s-orbitals). I've written about this in more detail at What is the exact definition of the radial distribution function?, so won't repeat myself here.
