[Solution] Given $\pi(R+x)^{2} H=\pi R^{2}(H+x)$, i.e., $H x^{2}+\left(2 R H-R^{2}\right) x=0$. Since $x \neq 0$, we have $H x+2 R H-R^{2}=0$. Therefore, $x=\frac{R^{2}-2 R H}{H}=\frac{8^{2}-2 \cdot 8 \cdot 3}{3}=\frac{16}{3}$. Hence, the answer is $(C)$.
$24 \cdot 2$ A cylinder has a radius of 8 inches and a height of 3 inches. The volume $V=\pi R^{2} H$. When $R$ increases by $x$ inches and $H$ increases by $x$ inches, the volume increases by the same fixed positive amount. What is the value of $x$ that satisfies this condition?
(A) No real value.
(B) Is an integer.
(C) Is a rational number, but not an integer.
(D) Is an irrational number.
(E) Is two real numbers.
(11th American High School Mathematics Examination, 1960)