b
Given $\frac{\mathrm{nv}}{2 \ell}=315$ and $(\mathrm{n}+1) \frac{\mathrm{v}}{2 \ell}=420$
$\Rightarrow \frac{n+1}{n}=\frac{420}{315} \Rightarrow n=3$
Hence $3 \times \frac{v}{2 \ell}=315 \Rightarrow \frac{v}{2 \ell}=105 \mathrm{Hz}$
Lowest resonant frequency is when $n=1$
Therefore lowest resonant frequency $=105 \mathrm{Hz}$