In general the electric potential (or electric field) at point P outside of a charge distribution can be expanded as a power series known as a multipole expansion : $$V(z)=\frac{A}{r}+\frac{B}{r^2}+\frac{C}{r^3}+\frac{D}{r^4}+...$$ where the coefficients $A, B, C, D, ...$ depend on the spherical polar and azimuthal co-ordinates $\theta, \phi$ and $r$ is the distance of P from centre O of the charge distribution. See Electric field far away from a bunch of charges.
In this case the total charge on the sphere is zero : $$Q=\int_0^R \int_0^{\pi} \rho r^2\sin\theta d\theta dr=KR\int_0^R (R-2r) dr \int_0^{\pi}\sin^2\theta d\theta=0$$ because the integral wrt $r$ is zero. So the electric potential far from the centre of the sphere will not be proportional to $1/r$ as it is for a point charge : that is, $A=0$ here.
It is convenient to divide the sphere into rings such as A and B which are parallel to the xy plane, because such rings have uniform charge density. A typical ring has radius $y=r\sin\theta$ and each point on it is the same distance $s$ from the point of interest P which lies on the $z$ axis. The potential at P due to the ring is simply $V=kQ/s$ where $k$ is Coulomb's constant,
Using the Cosine Rule for ring A $$s^2=z^2-2rz\cos\theta+r^2=z^2(1-x)$$ where $x=p(c-p), c=2\cos\theta, p=\frac{r}{z}$. The charge on each ring is $$dQ=2\pi y \rho r d\theta dr=2\pi\rho r^2\sin\theta d\theta dr$$
Ring B has the same charge as ring A and is located symmetrically about O, the centre of the sphere. The potential at P due to both rings A and B, each with charge $dQ$, is $$dV=\frac{dQ}{4\pi \epsilon_0 }(\frac{1}{s_1}+\frac{1}{s_2})=\frac{KR(R-2r)}{2\epsilon_0 z}(\frac{1}{\sqrt{1- x}}+\frac{1}{\sqrt{1+ x}})\sin^2\theta d\theta dr$$
The reciprocals can be expanded as power series : $$\frac{1}{\sqrt{1-x}}=1+\frac12x+\frac38x^2+\frac{5}{16}x^3+\frac{35}{128}x^4+\frac{63}{256}x^5+\frac{231}{1024}x^6+...$$ $$\frac{1}{\sqrt{1+x}}=1-\frac12x+\frac38x^2-\frac{5}{16}x^3+\frac{35}{128}x^4-\frac{63}{256}x^5+\frac{231}{1024}x^6+...$$ which have the half-sum $$\frac12 (\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1+x}})=1+\frac38x^2+\frac{35}{128}x^4++\frac{231}{1024}x^6+...$$
The advantage of taking 2 rings together is that we have halved the number of terms in the series. This simplifies calculation. To avoid double-counting we must reduce the range of integration to $\theta=0 \to \pi/2$.
The words "far from the centre" suggest that we should assume $z\gg R$ and therefore we ignore all except the lowest power terms. Substituting from above $$x^2=p^2(c-p)^2\approx \frac{r^2}{z^2}\cos^2\theta$$
The potential at P is approximately $$V\approx \frac{KR}{\epsilon_0 z} \int_0^{\pi/2} d\theta \int_0^R dr . (R-2r)(1+\frac38 \frac{r^2}{z^2} \cos^2 \theta ) \sin^2\theta$$ $$= \frac{KR}{\epsilon_0 z} \int_0^R (R-2r)\frac38 \frac{r^2}{z^2} dr \int_0^{\pi/2} \cos^2 \theta \sin^2\theta d\theta$$ $$= \frac{KR}{\epsilon_0 z} (-\frac{R^4}{16z^2} )(\frac{\pi}{16})$$ $$= -\frac{\pi KR^5}{256\epsilon_0 z^3}$$ The electric field at point P is $$E_z=-\frac{dV}{dz}=-\frac{3\pi KR^5}{256\epsilon_0 z^4}$$
Notes :
We should expect the potential to be -ve because the outer part of the sphere (which is closest to P) is negatively charged.
The dominant term in the charge distribution is not a dipole $V\propto 1/r^2$ but a quadrupole $V\propto 1/r^3$. We should expect to lose the dipole term because the centres of +ve and -ve charge coincide.
This result gives only the leading term in the multipole expansion and applies only for $z\gg R$. We could obtain more terms by retaining powers of $x^4, x^6...$ in the half-sum expansion and higher powers of $r$ when substituting for $x$. As more terms are retained the result becomes more accurate for smaller values of $z \gt R$.
