= Solution
Substitute $C=t^\alpha F(\xi)$, $\xi=rt^{-\beta}$. The derivatives give
$$
t^{\alpha-1}(\alpha F-\beta\xi F')
=Dt^{\alpha-2\beta}\left(F''+\frac2\xi F'\right).
$$
For a nontrivial spreading <similarity solution>, equality at all times requires $\boxed{\beta=1/2}$. The conserved total number of molecules is
$$
N=4\pi\int_0^\infty C(r,t)r^2\,dr
=4\pi t^{\alpha+3\beta}\int_0^\infty F(\xi)\xi^2\,d\xi.
$$
A finite nonzero constant $N$ therefore requires $\alpha+3\beta=0$, giving $\boxed{\alpha=-3/2}$. In one dimension the same diffusion length exponent is $1/2$, but conservation gives amplitude exponent $-1/2$; generally it is $-d/2$ in dimension $d$.
The profile equation is consequently
$$
\boxed{D\left(F''+\frac2\xi F'\right)+\frac12\xi F'+\frac32F=0.}
$$
Multiplication by $\xi^2$ puts it in the integrated form $[D\xi^2F'+\xi^3F/2]'=0$. For the regular localized <Gaussian heat kernel> profile, the constant is zero at the origin, so $DF'+\xi F/2=0$. Integrating gives
$$
\boxed{F(\xi)=A e^{-\xi^2/(4D)},\qquad
C(r,t)=\frac{N}{(4\pi Dt)^{3/2}}e^{-r^2/(4Dt)}.}
$$
The second expression chooses the optional normalization $A=N/(4\pi D)^{3/2}$. Its mass is $N$, and its width tends to zero with $t^{1/2}$ while its mass stays fixed, so it approaches $N$ times a <Dirac delta distribution> at the origin as $t\downarrow0$.
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