0 is one of the zeroes of $3x^2-7x+k$.
Use the product of the zeroes to find $k$, and then state the other zero.

The product of the zeroes is $\frac{c}{a}=\frac{k}{3}$. One zero is 0, so the product is 0 and $k=0$. The polynomial becomes $3x^2-7x=x(3x-7)$, whose second zero satisfies $3x=7$, i.e. $x=\frac{7}{3}$. (Sum check: $0+\frac{7}{3}=\frac{7}{3}=-\frac{b}{a}$.) The option $\frac{3}{7}$ inverts the fraction; $k=7$ confuses the constant with the middle coefficient; $k=3$ copies the leading coefficient.