Let $A(t)$ be the area of the square at time $t$, let $\ell(t)$ be the length of the side of the square at time $t$.
We are given the following information: the rate at which the length of the side (that is, $\ell(t)$) changes is 6 units per second. This is a way of giving you information about the derivative of $\ell$. That is, it's telling you that the rate of change of $\ell$ with respect to time (why time? because it's "per second") is $6$ units per second. In symbols, it's telling you that"
$$\frac{d\ell}{dt} = 6\text{ units/second}$$
We are asked to find how fast the area is changing when the area is 16 square units. "How fast" is a dead giveaway that they are asking you for the rate of change of the area; that is, they are asking you for the value of $\frac{dA}{dt}$, the derivative of the area, when the area is equal to 16. In symbols, they are asking you to find:
$$\frac{dA}{dt}\Bigm|_{A=16}.$$
The first step, as in any related rates problem, is to find an equation that relates the quantities involved, in this case the length of the side $\ell$ and the area $A$. The relation between $A$ and $\ell$ is given by:
$$A = \ell^2.$$
(Geometry: the area of a square is the square of the length of the side)
Once you have the relation between the original quantities, you take derivatives (implicitly) with respect to $t$ to obtain a relation between the quantities and their rates of change:
$$\begin{align*}
A &= \ell^2\\
\frac{d}{dt} (A) &= \frac{d}{dt}(\ell^2)\\
\frac{dA}{dt} &= 2\ell\frac{d\ell}{dt}.
\end{align*}$$
Once you have a relation between the quantities and the rates of change, you plug in all the information you have and solve for the rate you want.
Since we want to know $\frac{dA}{dt}$, we need to know the values of $\ell$ and of $\frac{d\ell}{dt}$. The value of $\frac{d\ell}{dt}$ we already know: they told us from the beginning that it was always equal to $6$. So we just need to figure out the value of $\ell$ at the instant we are interested in.
The instant we are interested in is the instant when $A=16$. Since $A=\ell^2$, we need $16=\ell^2$; solving for $\ell$ and remembering that it is a length, so it must be nonnegative, we find that $16=\ell^2$ means that $\ell$ must be equal to $4$. Plugging in everything (the value of $\ell$, the value of $\frac{d\ell}{dt}$) into the formula we got that relates everything we get
$$\begin{align*}
\frac{dA}{dt} &=2\ell\frac{d\ell}{dt}\\
\frac{dA}{dt} &= 2(4)(6)\\
&= 48\text{ units}^2/\text{second}.
\end{align*}$$
So at the instant that the area is 16 square units, the area is changing at a rate of 48 square units per second.
Best Answer
pedja's answer does seem to be expressed in a somewhat complicated way.
Let $A$ be the area in square centimeters. Let $s$ be the length of the side in centimeters. Let $t$ be time in seconds.
Then we are given $\dfrac{ds}{dt} = 6$.
We recall that $A = s^2$.
We want $\dfrac{dA}{dt}$ when $A=16$.
$$ \frac{dA}{dt} = \frac{d}{dt} s^2 = 2s \frac{ds}{dt}. $$
When $A=16$ then $s=4$ and $ds/dt = 6$. So $$ 2s\frac{ds}{dt} = 2\cdot4\cdot 6. $$