Answer
$27\text{ m}^3\text{ per meter}$
Work Step by Step
Let us say that the tangent Line contains the points $(x, c)$. The slope the tangent Line to the graph of $f(x)$ at $(x, c)$ can be written as $f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c} ...(1)$
In order to find the rate of change of volume with respect to radius, we will use equation (1).
Plug in the given data to obtain:
$$V'(3) =\lim\limits_{x \to 3} \dfrac{V(x)-V(3)}{(x-3)} \\=\lim\limits_{x \to 3} \dfrac{x^3-3^3}{(x-3)} \\= \lim\limits_{x \to 3} \dfrac{x^3-27}{x-3} \\=\lim\limits_{x \to 3} \dfrac{(x-3) (x^2+3x+9)}{(x-3)}\\=\lim\limits_{x \to 3} x^2+3x+9 \\=(3)^2+3(3)+9 \\=27 \ m^3/m$$
