We have \begin{equation*} F(x_0)=\infty \tag{1}\label{1} \end{equation*} for any nonzero $x_0$.

Indeed, by spherical symmetry, without loss of generality \begin{equation*} x_0=(2a,0,\dots,0) \end{equation*} for some real $a>0$.

Let $b_t:=B_{2a}((t,0,\dots,0))$ for real $t>0$. Let \begin{equation*} C_{d,a}:=\nu_{d-1}\big(B^{d-1}_a\big), \end{equation*} where $\nu_k$ is the standard Gaussian measure over $\Bbb R^k$ and $B^k_r$ is the ball in $\Bbb R^k$ of radius $r$ centered at the origin. Note that $b_t\supset[t-a,t-a/2]\times B^{d-1}_a$, if $\Bbb R^d$ is identified with $\Bbb R\times\Bbb R^{d-1}$. So, \begin{equation*} \nu(b_t)\ge\frac1{\sqrt{2\pi}}\int_{t-a}^{t-a/2} dx_1\,e^{-x_1^2/2} \;C_{d,a} \ge c_d\, e^{-(t-a/2)^2/2}, \end{equation*} where $c_d:=\frac{C_{d,a}}{\sqrt{2\pi}}\frac a2>0$.

On the other hand, \begin{equation*} \nu(b_t+x_0)\le\nu([t+2a-2a,\infty)\times\Bbb R^{d-1})= \frac1{\sqrt{2\pi}}\int_t^\infty dx_1\,e^{-x_1^2/2}\le e^{-t^2/2}. \end{equation*}

So, \begin{equation*} \frac{\nu(b_t)}{\nu(b_t+x_0)}\ge c_d\,\frac{e^{-(t-a)^2/2}}{e^{-t^2/2}} \to\infty \end{equation*} as $t\to\infty$. So, \eqref{1} is proved. $\quad\Box$

We have \begin{equation*} F(x_0)=\infty \tag{1}\label{1} \end{equation*} for any nonzero $x_0$.

Indeed, by spherical symmetry, without loss of generality \begin{equation*} x_0=(2a,0,\dots,0) \end{equation*} for some real $a>0$.

Let $b_t:=B_{2a}((t,0,\dots,0))$ for real $t>0$. Let \begin{equation*} C_{d,a}:=\nu_{d-1}\big(B^{d-1}_a\big), \end{equation*} where $\nu_k$ is the standard Gaussian measure over $\Bbb R^k$ and $B^k_r$ is the ball in $\Bbb R^k$ of radius $r$ centered at the origin. Note that $b_t\supset[t-a,t-a/2]\times B^{d-1}_a$, if $\Bbb R^d$ is identified with $\Bbb R\times\Bbb R^{d-1}$. So, if $t>a$, then \begin{equation*} \nu(b_t)\ge\frac1{\sqrt{2\pi}}\int_{t-a}^{t-a/2} dx_1\,e^{-x_1^2/2} \;C_{d,a} \ge c_d\, e^{-(t-a/2)^2/2}, \end{equation*} where $c_d:=\frac{C_{d,a}}{\sqrt{2\pi}}\frac a2>0$.

On the other hand, \begin{equation*} \nu(b_t+x_0)\le\nu([t+2a-2a,\infty)\times\Bbb R^{d-1})= \frac1{\sqrt{2\pi}}\int_t^\infty dx_1\,e^{-x_1^2/2}\le e^{-t^2/2}. \end{equation*}

So, if $t>a$, then \begin{equation*} \frac{\nu(b_t)}{\nu(b_t+x_0)}\ge c_d\,\frac{e^{-(t-a/2)^2/2}}{e^{-t^2/2}} \to\infty \end{equation*} as $t\to\infty$. So, \eqref{1} is proved. $\quad\Box$

We have \begin{equation} F(x_0)=\infty \tag{1}\label{1} \end{equation} for any nonzero $x_0$.

Indeed, by spherical symmetry, without loss of generality \begin{equation} x_0=(2a,0,\dots,0) \end{equation} for some real $a>0$.

Let $b_t:=B_{2a}((t,0,\dots,0))$ for real $t>0$. Let \begin{equation} C_{d,a}:=\nu_{d-1}\big(B^{d-1}_a\big), \end{equation} where $\nu_k$ is the standard Gaussian measure over $\Bbb R^k$ and $B^k_r$ is the ball in $\Bbb R^k$ of radius $r$ centered at the origin. Note that $b_t\supset[t-a,t+a]\times B^{d-1}_a$, if $\Bbb R^d$ is identified with $\Bbb R\times\Bbb R^{d-1}$. So, \begin{equation} \nu(b_t)\ge\frac1{\sqrt{2\pi}}\int_{t-a}^{t+a} dx_1\,e^{-x_1^2/2} \;C_{d,a} \ge c_d \frac{e^{-(t-a)^2/2}}{t} \end{equation} eventually (that is, for all large enough $t>0$), where $c_d:=\frac{C_{d,a}}{2\sqrt{2\pi}}>0$.

On the other hand, \begin{equation} \nu(b_t+x_0)\le\frac1{\sqrt{2\pi}}\int_{t+2a-2a}^\infty dx_1\,e^{-x_1^2/2}\le e^{-t^2/2}. \end{equation}

So, eventually, \begin{equation} \frac{\nu(b_t)}{\nu(b_t+x_0)}\ge\frac{c_d}t\frac{e^{-(t-a)^2/2}}{e^{-t^2/2}} \to\infty \end{equation} as $t\to\infty$. So, \eqref{1} is proved. $\quad\Box$

We have \begin{equation*} F(x_0)=\infty \tag{1}\label{1} \end{equation*} for any nonzero $x_0$.

Indeed, by spherical symmetry, without loss of generality \begin{equation*} x_0=(2a,0,\dots,0) \end{equation*} for some real $a>0$.

Let $b_t:=B_{2a}((t,0,\dots,0))$ for real $t>0$. Let \begin{equation*} C_{d,a}:=\nu_{d-1}\big(B^{d-1}_a\big), \end{equation*} where $\nu_k$ is the standard Gaussian measure over $\Bbb R^k$ and $B^k_r$ is the ball in $\Bbb R^k$ of radius $r$ centered at the origin. Note that $b_t\supset[t-a,t-a/2]\times B^{d-1}_a$, if $\Bbb R^d$ is identified with $\Bbb R\times\Bbb R^{d-1}$. So, \begin{equation*} \nu(b_t)\ge\frac1{\sqrt{2\pi}}\int_{t-a}^{t-a/2} dx_1\,e^{-x_1^2/2} \;C_{d,a} \ge c_d\, e^{-(t-a/2)^2/2}, \end{equation*} where $c_d:=\frac{C_{d,a}}{\sqrt{2\pi}}\frac a2>0$.

On the other hand, \begin{equation*} \nu(b_t+x_0)\le\nu([t+2a-2a,\infty)\times\Bbb R^{d-1})= \frac1{\sqrt{2\pi}}\int_t^\infty dx_1\,e^{-x_1^2/2}\le e^{-t^2/2}. \end{equation*}

So, \begin{equation*} \frac{\nu(b_t)}{\nu(b_t+x_0)}\ge c_d\,\frac{e^{-(t-a)^2/2}}{e^{-t^2/2}} \to\infty \end{equation*} as $t\to\infty$. So, \eqref{1} is proved. $\quad\Box$