= Solution
Uniform probability per volume does not give a uniform radial distribution. A shell of radius $r$ and thickness $dr$ occupies volume $4\pi r^2dr$, so the radial <probability density function> is
$$
f(r)=\frac{3r^2}{R^3},\qquad0\leq r\leq R.
$$
It integrates to one. The second radial moment is
$$
\langle r^2\rangle=\int_0^R r^2\frac{3r^2}{R^3}dr=\frac35R^2,
\qquad\boxed{r_{\rm rms}=\sqrt{\frac35}R=1.162\ \mathrm{fm}\quad(R=1.5\ \mathrm{fm}).}
$$
The <radial uncertainty scale in a uniform sphere> supplies the position-uncertainty scale \b[as instructed by the toy model]. It is not literally the <standard deviation> of the scalar radius: $\langle r\rangle=3R/4$ and $\sqrt{\langle r^2\rangle-\langle r\rangle^2}=R\sqrt{3/80}$. Nor is a uniform spatial probability density sufficient to determine a unique <momentum space> <wavefunction>. This explicit distinction keeps the order-of-magnitude assumption separate from a fully specified quantum bound-state calculation. Replacing the radial density by a uniform density in $r$ would give a different second moment because it omits the spherical volume element.
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