The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius = $$\sqrt{3}$$ is:

Let the given sphere have radius $$R=\sqrt{3}\;.$$ We wish to fit inside it a right circular cylinder having base-radius $$r$$ and height $$h$$ in such a way that the volume is as large as possible.

The axis of the “right” cylinder will pass through the centre of the sphere, so if we measure upward from the centre, the top rim of the cylinder has coordinates $$\bigl(r,\,y\bigr)$$ in a suitable coordinate system. This point must lie on the sphere, therefore we have the relation of the sphere:

$$r^{\,2}+y^{\,2}=R^{\,2}\;.$$

The total height of the cylinder is twice the half-height $$y$$, that is

$$h = 2y\;.$$

Now we write the formula for the volume of a right circular cylinder:

Volume $$V=\pi \,(\text{base-radius})^{2}\times(\text{height})\;.$$

So, substituting the symbols we are using,

$$V = \pi r^{\,2}\,h = \pi r^{\,2}\,(2y)=2\pi r^{\,2}y\;.$$

Because $$y$$ is related to $$r$$ through the sphere equation, we solve that equation for $$y$$:

$$y^{\,2}=R^{\,2}-r^{\,2}\;\;\Longrightarrow\;\; y=\sqrt{R^{\,2}-r^{\,2}}\;.$$

Now we can express the cylinder’s volume as a function of only one independent variable, namely $$r$$:

$$V(r)=2\pi r^{\,2}\sqrt{R^{\,2}-r^{\,2}}\;.$$

To obtain the maximum volume we differentiate with respect to $$r$$ and set the derivative equal to zero.

First we compute the derivative. Write $$V(r)=2\pi\,f(r)$$ with $$f(r)=r^{\,2}(R^{\,2}-r^{\,2})^{1/2}\;.$$ Using the product rule and the chain rule:

$$\dfrac{d}{dr}f(r)=2r\bigl(R^{\,2}-r^{\,2}\bigr)^{1/2}+r^{\,2}\cdot\dfrac{1}{2}\bigl(R^{\,2}-r^{\,2}\bigr)^{-1/2}\cdot(-2r)\;.$$

Simplifying inside the brackets:

$$\dfrac{d}{dr}f(r)=2r(R^{\,2}-r^{\,2})^{1/2}-\dfrac{r^{\,3}}{(R^{\,2}-r^{\,2})^{1/2}}\;.$$

Therefore

$$\dfrac{dV}{dr}=2\pi\left[\,2r(R^{\,2}-r^{\,2})^{1/2}-\dfrac{r^{\,3}}{(R^{\,2}-r^{\,2})^{1/2}}\,\right]\;.$$

For a maximum or minimum we set this derivative equal to zero. The factor $$2\pi$$ and the denominator $$(R^{\,2}-r^{\,2})^{1/2}$$ are never zero within the sphere, so we concentrate on the numerator:

$$2r(R^{\,2}-r^{\,2})-r^{\,3}=0\;.$$

Now we solve step by step:

$$2r(R^{\,2}-r^{\,2})-r^{\,3}=0$$

$$\Longrightarrow\; 2rR^{\,2}-2r^{\,3}-r^{\,3}=0$$

$$\Longrightarrow\;2rR^{\,2}-3r^{\,3}=0$$

$$\Longrightarrow\; r\bigl(2R^{\,2}-3r^{\,2}\bigr)=0\;.$$

The factor $$r=0$$ would correspond to no cylinder at all, so we take the other factor equal to zero:

$$2R^{\,2}-3r^{\,2}=0$$

$$\Longrightarrow\; r^{\,2}=\dfrac{2}{3}R^{\,2}$$

$$\Longrightarrow\; r=R\sqrt{\dfrac{2}{3}}\;.$$

Next we find $$y$$ and then $$h$$. From $$y=\sqrt{R^{\,2}-r^{\,2}}$$ we have

$$y=\sqrt{R^{\,2}-\dfrac{2}{3}R^{\,2}}=\sqrt{\dfrac{1}{3}R^{\,2}}=\dfrac{R}{\sqrt{3}}\;.$$

So the optimal cylinder’s height is

$$h = 2y = 2\cdot\dfrac{R}{\sqrt{3}}=\dfrac{2R}{\sqrt{3}}\;.$$

Now substitute these optimal dimensions back into the volume formula:

$$V_{\text{max}}=\pi r^{\,2}h =\pi\left(\dfrac{2}{3}R^{\,2}\right)\left(\dfrac{2R}{\sqrt{3}}\right) =\pi\cdot\dfrac{2}{3}\cdot\dfrac{2}{\sqrt{3}}\,R^{\,3} =\dfrac{4\pi R^{\,3}}{3\sqrt{3}}\;.$$

Finally we put the numerical value $$R=\sqrt{3}$$ into this expression.

Because $$R^{\,3}=(\sqrt{3})^{3}=3\sqrt{3}\;,$$ we get

$$V_{\text{max}}=\dfrac{4\pi\,(3\sqrt{3})}{3\sqrt{3}}=4\pi\;.$$

Thus the largest possible right circular cylinder that can be inscribed in the given sphere has volume $$4\pi\;.$$

Hence, the correct answer is Option C.