$\newcommand\si\sigma\newcommand\R{\Bbb R}$Let $f$ be the p.d.f. of the normal distribution $N(\mu,\si^2I_n)$ ovr $\R^n$, where $I_n$ is the identity matrix. Then for all $x\in\R^n$ $$f(x)=\int_{(0,\infty)}dt\,1(t<f(x))=\int_{(0,\infty)}dt\,1(x\in B_t) \\ =\int_{(0,\si^{-n})}dt\,|B_t|\frac{1(x\in B_t)}{|B_t|} =\int_{(0,\si^{-n})}dt\,p(t)f_t(x),$$ where $B_t:=\{x\in\R^n\colon f(x)>t\}$ -- which is a nonempty open ball in $\R^n$ centered at $\mu$ if $t\in(0,\si^{-n})$, $|B_t|$ is the Lebesgue measure of $B_t$, $p(t):=|B_t|$, and $f_t$ is the p.d.f. of the uniform distribution over the ball $B_t$.
Thus, $N(\mu,\si^2I_n)$ is the mixture of the uniform distributions over the balls $B_t$, with the mixing p.d.f. $p$ supported on the interval $(0,\si^{-n})$.
(That $p$ is a p.d.f. follows because $$1=\int_{\R^n}dx\,f(x)=\int_{(0,\si^{-n})}dt\,p(t)\int_{\R^n}dx\,f_t(x) =\int_{(0,\si^{-n})}dt\,p(t).\ )$$