You'll be all alone!

It is rather easy, actually. First, by Hubble's relation we have $V = Ha$ where $V$ is the velocity of a galaxy that is $a$ meters away from us. To know the distance the galaxy will be from us in the moment it will desapear, we make $V = c$ (it is allowed in general relativity). Again, using Hubble's relation

$\frac{da}{dt} = Ha \Rightarrow \int_{x_{0}}^{x} \frac{da}{a} = H \int_{t_{0}}^{t} dt \therefore x = x_{0}e^{H(t-t_{0})}$

Where $x_{0}$ is the distance of the nearest galaxy to us, $x$ is the distance function and we'll assume $t_{0} = 0$ (that is, the time starts counting from now). Using the condition we discovered above (that the distance must be $a = \frac{c}{H}$ ) we have

$\frac{c}{H} = x_{0}e^{Ht} \Rightarrow t = \frac{1}{H} \times \ln{\frac{c}{x_{0}H}}$

And for those values that were given in the problem, we'll get the time in seconds. Converting to years, the answer will give us $n = 11$ .