if the surface area of a cube is increasing at a rate of $3\cdot 6{\mathrm{cm}}^2/\sec$, retaining ifs shape, then the rate of change of its volume (in $c{m}^3$ / sec), when the length of a side of the cube is $10 cm$, is 

Let the length an edge of the cube at any moment $t$

be $x cm.$ Then, its surface area $S$ and volume $V$ are given by 

$S=6x^2$ and $V=x^3$

$\Rightarrow \frac{dS}{dt}=12x\frac{dx}{dt}$ and $\frac{dV}{dt}=3x^2\frac{dx}{dt}$

$\Rightarrow 3\cdot 6=12\times 10\frac{dx}{dt}$ and $\frac{dV}{dt}=3\times {10}^2\times \frac{dx}{dt}$

$\begin{array}{l}\Rightarrow \frac{dx}{dt}=\frac{3}{100}\text{ and }\frac{dV}{dt}=300\frac{dx}{dt}\\ \Rightarrow \frac{dV}{dt}=300\times \frac{3}{100}=9.\end{array}$