In this problem, in all cases, you are expected to evaluate any integrals in your answers.
Along the axis of symmetry, \(s=0\), and the integral becomes \begin{align} V(z) &= \frac{1}{4\pi\epsilon_0} \int_0^{2\pi}\frac{\lambda\,R\,d\phi'}{\sqrt{R^2 + z^2}}\\ &= \frac{1}{4\pi\epsilon_0} \frac{2\pi\lambda\,R}{\sqrt{R^2 + z^2}}\\ &= \frac{1}{4\pi\epsilon_0} \frac{Q}{\sqrt{R^2 + z^2}} \end{align} Sense-making: Always check dimensions on these potential expressions---especially because electric fields are coming up, and they have different dimensions. Also, note that there's an interesting symmetry between \(R\) and \(z\) here: moving a small distance away from the ring along the \(z\)-axis has the same effect as staying put while making the ring a little bit bigger.
The disk is a 2D object, so we need a \(\sigma\) to describe it, thus our tiny amount of charge takes on the form: \begin{equation} \sigma dA = \sigma\, s' \,ds' d\phi' \end{equation} So we use this in our general expression for \(V\): \begin{align} V(z) &= \frac{1}{4\pi\epsilon_0}\int \frac{\sigma(r') dA'}{|\vec{r}-\vec{r}'|} \\ &= \frac{\sigma}{4\pi\epsilon_0}\int_0^{2\pi} \int_0^R \frac{s'\,ds' \, d\phi'}{\sqrt{s'^2 + z^2}} \\ &= \frac{2\pi\sigma}{4\pi\epsilon_0}\int_0^R \frac{ s'\,ds'}{\sqrt{s'^2 + z^2}} \\ &= \frac{2\pi\sigma}{4\pi\epsilon_0} \sqrt{s'^2 + z^2} \Bigg|_0^R \\ &= \frac{2\pi\sigma}{4\pi\epsilon_0} \left( \sqrt{R^2 + z^2} - |z| \right) \end{align}
Sense-making: Pay close attention to the different dimensions of \(\sigma\), and you might even want to put the whole expression back in terms of \(Q\) like we did for the previous part. Also, at first glance it seems like this potential might not be falling off as \(z\) gets larger, which wouldn't make sense at all. You might want to graph the potential, or even graph the two terms in the potential separately, to see that the first term may start off higher, but that it quickly approaches the second term in value, while always remaining positive.
If \(\vert z\vert\gg R\), then \begin{align} \sqrt{R^2 + z^2} &= |z| \sqrt{1+\frac{R^2}{z^2}}\\ &= |z| \left( 1 + \frac{1}{2} \frac{R^2}{z^2} - \frac{1}{8} \frac{R^4}{z^4} + ... \right) \end{align} Inserting this into the result obtained in (b) leads to \begin{align} V(z) &= \frac{2\pi\sigma}{4\pi\epsilon_0} \left( \sqrt{R^2 + z^2} - |z| \right)\\ &= \frac{1}{4\pi\epsilon_0}\frac{\sigma \pi R^2}{|z|} \left( 1 - \frac{R^2}{4z^2} + ... \right) \end{align}
Sense-making: Aha, our old friend \(1/z\) dependence shows up clearly here, which makes sense because we have a nonzero total charge (this is the “monopole” term). However, we know it isn't really a point charge, so the fact that the term of next leading order is negative corrects the potential downward as a result of the charge being somewhat spread out on the disc. Also note the absolute value on \(z\). Based on the symmetry of the charge distribution, we expect the potential to be described by an even function. \(\frac{1}{z}-\frac{1}{z^3}\) is an odd function, but \(\frac{1}{|z|} - \frac{1}{|z|^3}\) is an even function.
Also, I pulled out the overall factor of \(\sigma 2\pi R^2\) since that is the total charge \(Q\) on the disk.
Interestingly, if you power series expand the answer to part a and put them both in terms of \(Q\), you get: \begin{align} V_{\textrm{ring}} &\approx \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{|z|} - \frac{R^2}{2|z|^3} \right)\\ V_{\textrm{disc}} &\approx \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{|z|} - \frac{R^2}{4|z|^3} \right) \end{align} Very far away, i.e. to leading order, the solutions look exactly the same, but the next order term shows that the potential due to the disc doesn't fall off as fast as the ring (that second nonzero term is negative, so it's taking away less from \(V_{disc}\) than it is from \(V_{ring}\).