Answer
Rhoda should compress the spring a distance of $~~1.25~cm$
Work Step by Step
The time of flight $t$ is equal in both cases.
We can find an expression for the speed $v_B$ of the marble as it leaves the spring when Bobby compresses the spring:
$x = v_B~t$
$v_B = \frac{x}{t}$
$v_B = \frac{1.93~m}{t}$
We can find an expression for the required speed $v_R$ of the marble as it leaves the spring when Rhoda compresses the spring:
$D = v_R~t$
$v_R = \frac{D}{t}$
$v_R = \frac{2.20~m}{t}$
We can divide $v_R$ by $v_B$ to find the required speed $v_R$ in terms of $v_B$:
$\frac{v_R}{v_B} = \frac{\frac{2.20~m}{t}}{\frac{1.93~m}{t}}$
$\frac{v_R}{v_B} = \frac{2.20}{1.93}$
$v_R = \frac{2.20}{1.93}~v_B$
We can find an expression for the compression distance $d_B$ when Bobby compresses the spring:
$\frac{1}{2}mv_B^2 = \frac{1}{2}kd_B^2$
$d_B^2 = \frac{m}{k}v_B^2$
$d_B = \sqrt{\frac{m}{k}}~v_B$
We can find the required compression distance $d_R$ when Rhoda compresses the spring:
$\frac{1}{2}mv_R^2 = \frac{1}{2}kd_R^2$
$d_R^2 = \frac{m}{k}v_R^2$
$d_R = \sqrt{\frac{m}{k}}~v_R$
$d_R = \frac{2.20}{1.93}~\sqrt{\frac{m}{k}}~v_B$
$d_R = \frac{2.20}{1.93}~d_B$
$d_R = (\frac{2.20}{1.93})~(1.10~cm)$
$d_R = 1.25~cm$
Rhoda should compress the spring a distance of $~~1.25~cm$
