Answer
$90\;\rm N/m$
Work Step by Step
The tension in the bungee cord is equal to the spring force [Recall Hooke's law].
$$T_s=kx\tag 1$$
where $k$ is the spring constant, and $x$ is the stretched distance.
We also know that the tension in the cord when it is vibrating and producing standing waves is given by
$$v=\sqrt{\dfrac{T_s}{\mu}}$$
where $v$ is the speed of the wave and $\mu$ is the linear density of the cord.
Recall that $v=\lambda f$, so
$$\lambda f=\sqrt{\dfrac{T_s}{\mu}}$$
The wave in the cord with two antinodes, so $\lambda=L'$
where $L'$ is the total length of the cord when it is stretched while $L$ is its normal length.
$$L'f=\sqrt{\dfrac{T_s}{\mu}}$$
squaring both sides;
$${L'}^2f^2= \dfrac{T_s}{\mu} $$
Noting that $\mu=m_{cord}/L'$;
Thus,
$$\dfrac{{L'}^{ \color{red}{\bf\not} 2}f^2m_{cord}}{ \color{red}{\bf\not} L'}= T_s $$
Plugging into (1);
$${L'}f^2m_{cord}=kx $$
$$k=\dfrac{{L'}f^2m_{cord}}{x}$$
Plugging the known;
$$k=\dfrac{{(1.8)}(20)^2(0.075)}{(0.6)}$$
where $x=L'-L$
$$k=\color{red}{\bf 90}\;\rm N/m$$
