We do an analysis, and arrive after a while at a cubic equation. This equation is then best tackled using a numerical method.
First we look at the Physics of the situation. The weight of water displaced by the sphere is the same as the weight of the sphere, by a principle that goes back to Archimedes.
Units were not specified, but it seems clear that we are expected to work in grams per cc. Then water can be taken to have density $1$, though the actual density is (slightly) dependent on temperature and pressure. So the weight of the water displaced is, in the appropriate units, equal to $1$ times the volume of the spherical cap. This is equal to the weight of the sphere. You have been given a formula for that. In our case, $d=0.6$.
The spherical cap has volume
$$\frac{\pi}{3}(3rh^2-h^3).$$
So we want to solve the equation
$$\frac{\pi}{3}(3rh^2-h^3)(1)=\frac{4\pi r^3}{3}(0.6) .$$
We want to find $h$ as a fraction of the radius. Note that since the density of the ball is greater than $0.5$, more than half the sphere will be submerged, unlike what is suggested by the picture!
Let $h=tr$. Substitute for $h$ in the above equation. There is a lot of cancellation, and after we do that we reach
$$3t^2-t^3=(4)(0.6), \qquad\text{or equivalently}\qquad t^3-3t^2+2.4=0.$$
There is a formula for the roots of a cubic. However, it is not very useful in this situation. It is best to solve the equation numerically, using your favourite numerical procedure.
Let $f(t)=t^3-3t^2+2.4$. We know the root we are interested in is $>1$, but probably not by much. Note that $f(1.1)=0.101$ and $f(1.2)=-0.192$. So our root is between $1.1$ and $1.2$, probably closer to $1.1$. You should end up with something around $1.134$.
I do not know what method you are now expected to use. Maybe the Newton-Raphson method? If you have trouble setting that up, please leave a message.
We can apply the Intersecting Chords Theorem.
You chord length is the length $UV$ and the segment height is the length $PX$.
The intersecting chords theorem tells us that $XP \times XQ = XU \times XV$.
Let $\ell = UV$ and $h=XP$. It follows that $UX = XV = \tfrac{1}{2}\ell$. The ICT then tells us that
$$\tfrac{1}{2}\ell \times \tfrac{1}{2}\ell = h \times XQ \, ,$$
i.e. $XQ = \tfrac{1}{4h}\ell^2$. The diameter $PQ=PX+XQ$ and
$$PX + XQ = h + \frac{\ell^2}{4h}=\frac{4h^2+\ell^2}{4h}$$
The radius is then one half of this, i.e.
$$CQ = \frac{4h^2+\ell^2}{8h} \, . $$
Best Answer
With the notations in this figure (borrowed from https://en.wikipedia.org/wiki/Circular_segment), $c$ is your chord length $L$, $$\theta=2\arcsin(\frac{c}{2R}),$$ the green area is $\frac{R^2}{2}(\theta-\sin(\theta))$. The larger circular segment area (whole area of the disk minus the green area) is given by $$A=R^2(\pi-\theta/2+\sin(\theta)/2).$$ If you replace $\theta$ from the first equation, you obtain a transcendental equation in $R$. It is possible to solve this numerically.
Similar equations for a spherical cap can be found at https://en.wikipedia.org/wiki/Spherical_cap