= Solution
Work first on $B=B_1$, and let $m=|B|^{-1}\int_Bu$. Put $Z=\{u=0\}$. Since $|Z|\geq\theta|B|$,
$$
\theta|B|m^2\leq |Z|m^2=\int_Z|u-m|^2\leq\int_B|u-m|^2.
$$
Also $\int_Bu^2=\int_B|u-m|^2+|B|m^2$, so
$$
\int_Bu^2\leq(1+\theta^{-1})\int_B|u-m|^2.
$$
The <Poincare-Wirtinger inequality> now bounds the last integral by $C_n\int_B|Du|^2$. Here is a direct proof on the ball. For smooth $u$, the variance identity and the fundamental theorem along line segments give
$$
\int_B|u-m|^2=\frac1{2|B|}\int_B\int_B|u(x)-u(y)|^2\,dx\,dy,
$$
$$
|u(x)-u(y)|^2\leq4\int_0^1|Du((1-t)x+ty)|^2\,dt.
$$
For $0\leq t\leq1/2$, fix $y$ and change variables $z=(1-t)x+ty$. Convexity keeps $z$ inside $B$, and the inverse Jacobian is at most $2^n$. Thus the double integral of this <gradient> term is at most $2^n|B|\int_B|Du|^2$. For $1/2\leq t\leq1$, fix $x$ instead and obtain the same bound. Integration over $t$ proves $\int_B|u-m|^2\leq2^{n+1}\int_B|Du|^2$. Density of smooth functions in $W^{1,2}(B)$ passes the estimate to all Sobolev functions.
For a ball centered at $a$ with radius $R$, set $v(y)=u(a+Ry)$. Its zero set has at least the same volume fraction $\theta$, while $\int_{B_1}v^2=R^{-n}\int_{B_R}u^2$ and $\int_{B_1}|Dv|^2=R^{2-n}\int_{B_R}|Du|^2$. Therefore
$$
\boxed{\int_{B_R}u^2\leq2^{n+1}(1+\theta^{-1})R^2\int_{B_R}|Du|^2.}
$$
This establishes the <Poincare inequality with a positive-measure zero set>, with an explicit admissible constant. If $\theta=1$, the function is already zero <almost everywhere>.
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