Lebesgue measure of any line in $\mathbb{R^2}$.

If the line is $$ \ell=\{(x,0): x\in\mathbb R\}, $$ then for every $\varepsilon>0$, we have $$ \ell\subset \bigcup_{k\in\mathbb Z}I_k^\varepsilon, $$ where $$ I_k^\varepsilon=[k,k+1]\times[2^{-|k|-2}\varepsilon,-2^{-|k|-2}\varepsilon]. $$ But $$ m_2(\ell)\le\sum_{k\in\mathbb Z} m_2(I_k)=\sum_{k\in\mathbb Z} 2^{-|k|-1}\varepsilon=\varepsilon. $$ Thus $m_2(\ell)<\varepsilon$, for every $\varepsilon>0$, and hence $m_2(\ell)=0$.

Any other straight line in $\mathbb R^2$ is obtained by a rigid motion of $\ell$, and rigid motion does not change the Lebesgue measure of a set.

Indeed, every proper linear subspace (or more generally, every proper hyperplane) of $\mathbb R^n$ has zero $n-$dimensional Lebesgue measure.