Hint: Here differentiate the volume of sphere formula wrt to time and equate it to zero and substitute the given values in the question to find the rate of increase of outer radius.

Complete step by step answer:

Let $r_i$ and $r_o$ are the inner and outer radii of the hollow sphere respectively.

Given, $\dfrac{{d{r_i}}}{{dt}} = 1{\text{ cm/sec }} $

Volume of the hollow sphere is given by $V = \dfrac{4}{3}\pi \left( {{r_o}^3 - {r_i}^3} \right) $

Also, given that the volume of the hollow sphere is constant.

I.e. $ V = \dfrac{4}{3}\pi \left( {{r_o}^3 - {r_i}^3} \right) $= constant

Differentiating the above equation with respect to time, we get

$\dfrac{{d{\text{V}}}}{{dt}} = \dfrac{4}{3}\pi \left( {3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}}} \right) $

Since, differentiation of a constant is zero. Therefore, $\dfrac{{d{\text{V}}}}{{dt}} = 0$.

$ \Rightarrow 0 = \dfrac{4}{3}\pi \left( {3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}}} \right) \Rightarrow 0 = 3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2\dfrac{{d{r_i}}}{{dt}} $

Putting the value of $\dfrac{{d{r_i}}}{{dt}} = 1 cm/sec$, we get

$ \Rightarrow 0 = 3{r_o}^2\dfrac{{d{r_o}}}{{dt}} - 3{r_i}^2 \Rightarrow \dfrac{{d{r_o}}}{{dt}} = \dfrac{{{r_i}^2}}{{{r_o}^2}}{\text{ }}...........{\text{(1)}} $

Since, we have to find rate of increase of outer radii of the hollow sphere i.e.

$\dfrac{{d{r_o}}}{{dt}}$ when $r_i = 4$ cm and $r_o = 8$ cm

So put $r_i = 4$ cm and $r_o = 8$ cm in equation (1), we have

$\dfrac{{d{r_o}}}{{dt}} = \dfrac{{{4^2}}}{{{8^2}}} = \dfrac{1}{4}{\text{ cm/sec}}$.

Since, the above rate of change of inner radii is positive that means its increasing with time.

Note - These types of problems are always computed by somehow obtaining the relation between the rate of change of two quantities where the value of one is given and the value of the other is to be computed.