The integral
$$
\int_0^{\infty} A e^{-\rho^2/2\Delta^2} R(\rho) \rho \,d\rho \tag{1}
$$
is some number, call it $C$. It will be a common factor in all calculations since it does not depend on $m$.
Basically, because the radial part of the probability distribution does not depend on $m$, expand
$$
\cos(\varphi)^2=\left(e^{2i\varphi} +2 +e^{-2i\varphi}\right)\, . \tag{2}
$$
Using $e^{im\varphi}/{\sqrt{2\pi}}$ as normalized states over the circle, the probability of getting $m=0$ is related to the overlap
$$
\int_0^{\infty} A e^{-\rho^2/2\Delta^2} R(\rho) \rho \,d\rho
\times \int_0^{2\pi} d\varphi \cos(\varphi^2) \frac{1}{\sqrt{2\pi}}= C \sqrt{\frac{\pi}{2}}\, , \\
P(m=0)=\vert C\vert^2 \frac{\pi}{2}
$$
The probability of getting $m=\pm 2$ is likewise given by
$$
\int_0^{\infty} A e^{-\rho^2/2\Delta^2} R(\rho) \rho \,d\rho
\times \int_0^{2\pi} d\varphi \cos(\varphi^2) \frac{e^{\pm 2i\varphi}}{\sqrt{2\pi}}= \frac{C}{2} \sqrt{\frac{\pi}{2}}\, , \\
P(m=\pm 2)=\vert C\vert^2 \frac{1}{4}\frac{\pi}{2}\, .
$$
As the probabilities must sum to 1 you can now determine $\vert C\vert^2$:
$$
\vert C\vert^2\frac{\pi}{2}\left(\frac14 + 1+ \frac14\right)=1
\quad\Rightarrow\quad \vert C\vert^2=\frac{4}{3\pi}
$$
so
$$
P(m=0)=\frac{4}{3\pi}\times \frac{\pi}{2}= \frac{2}{3}\, ,\\
P(m=\pm 2)=\frac{4}{3\pi}\times \frac{\pi}{8}=\frac{1}{6}\, .
$$
As a shortcut, note that all the $\Phi_m$ states share the same radial function so the integral (1) will be common to all $m$-states; the square of the coefficients in (2) give now the unnormalized probabilities of getting each $m$ outcome. Since the sum of the coefficients is $6$, the probabilities are then
$$
m=\pm 2: 1/6\, ,\qquad m=0: 4/6=2/3
$$
of course as before.
