= Solution
The <zero extension of H01> makes $v_m$ an element of $H^1(\mathbb R^3)$. The <chain rule> and <change of variables> give the exact scaling identities
$$
\boxed{\|\nabla v_m\|_{L^2(\mathbb R^3)}^2=\|\nabla u_m\|_{L^2(B)}^2,\qquad
\|v_m\|_{L^2(\mathbb R^3)}^2=r_m^{-2}\|u_m\|_{L^2(B)}^2.}
$$
Indeed $\nabla v_m(x)=r_m^{3/2}\nabla u_m(r_mx)$ inside the expanded ball, and there is no boundary jump because the <Sobolev trace> is zero.
First take the printed convergence assertion as an assumption, as the question requests. It would imply $\|\nabla v_m\|_2\to\|\nabla v\|_2>0$. The strict inequality follows because an $H^1(\mathbb R^3)$ function with zero <weak gradient> is constant <almost everywhere>, and the only constant in $L^2(\mathbb R^3)$ is zero; this contradicts the asserted positivity of $v$. If $u_m$ converged strongly in $H^1(B)$, its limit would have to be its <weak limit> zero from (iii), forcing $\|\nabla u_m\|_2\to0$. The first scaling identity contradicts this. Thus the intended answer is
$$
\boxed{u_m\text{ does not converge strongly in }H^1(B).}
$$
There is nevertheless a substantive printing error: the asserted strong convergence of $v_m$ in the full $H^1(\mathbb R^3)$ is false for these positive sources. Here is a check that does not rely on any assertion about the limiting profile. Put $R_m=r_m^{-1}$ and $c=|E|^{-5/6}$. Rescaling the <weak formulation> gives
$$
-\Delta v_m=c\chi_E\quad\text{in }B_{R_m},\qquad v_m=0\quad\text{on }\partial B_{R_m}.
$$
The <spherical mean of a Poisson solution outside its source> provides an exact obstruction. The total source is $M=c|E|=|E|^{1/6}>0$. Let $a_m(s)=(4\pi)^{-1}\int_{S^2}v_m(s\omega)\,d\omega$ be its <spherical mean>. On $1<s<R_m$ the function is a <harmonic function>. The radial flux identity, obtained by testing against radial <test functions> or by the <divergence theorem>, gives $-4\pi s^2a_m'(s)=M$, while the zero boundary value gives $a_m(R_m)=0$. Hence
$$
a_m(s)=\frac{M}{4\pi}\left(\frac1s-\frac1{R_m}\right)\qquad(1<s<R_m).
$$
<Cauchy-Schwarz inequality> on each sphere, followed by <spherical coordinates>, now yields
$$
\begin{aligned}
\|v_m\|_2^2
&\geq4\pi\int_1^{R_m}s^2a_m(s)^2\,ds\\
&=\frac{M^2}{4\pi}\left(\frac{R_m}{3}-1+\frac1{R_m}-\frac1{3R_m^2}\right)
\longrightarrow\infty.
\end{aligned}
$$
Thus the printed sequence is not even bounded in the full $H^1$ <norm>. The result on <Poisson convergence on expanding balls in a homogeneous Sobolev space> gives the appropriate corrected convergence in the <homogeneous Sobolev space> $\dot H^1(\mathbb R^3)$, which controls the <gradient> in $L^2$ and the function in $L^6$, without requiring it to belong to $L^2$.
For completeness, the desired nonconvergence of $u_m$ has a direct proof that needs neither the incorrect claim nor its correction. Choose a nonnegative nonzero <test function> $\eta\in C_c^\infty(E)$, which exists since $E$ is a domain containing the origin. The rescaled <weak formulation> and <Cauchy-Schwarz inequality> give
$$
0<c\int_E\eta\,dx
=\int_{\mathbb R^3}\nabla v_m\cdot\nabla\eta\,dx
\leq\|\nabla v_m\|_2\|\nabla\eta\|_2.
$$
Consequently, for every $m$,
$$
\boxed{\|\nabla u_m\|_2=\|\nabla v_m\|_2
\geq\frac{|E|^{-5/6}\int_E\eta\,dx}{\|\nabla\eta\|_2}>0.}
$$
Together with (iii), this rigorously proves failure of <strong convergence> for the actual sequence defined in the paper.
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