This is a quadratics problem. The full question reads:
An open container with a square base is made by cutting $4~\text{cm}$ square pieces out of a piece of tin. If the volume of the container is $196~\text{cm}^3$, find the dimensions of the original template. Let the length of the tinplate be $x~\text{cm}$.
I surmise that $x-8~\text{cm}$ is the unspecified length, but I'm unsure about anything else.
Make the box by cutting off the corners of the $x$ cm $\times x$ cm "original template" and folding up the sides:
Box base area $=\frac{\text{volume}}{\text{height}}=\frac{196}{4} = 49 \text{ cm}^2$
$b = \sqrt{49} = 7$ cm
$x= b+8 = 15 $ cm
Alternative answer if the box was not specified as square. Then the piece of tin that was used is $x$ cm wide and $y$ cm long, and
\begin{align}(x-8)(y-8) &= 49 \\[2ex] xy-8x-8y &= 49-64 = -15 \\[2ex] (x-8)y &=8x-15\\[2ex] y &=\frac{8x-15}{x-8} \end{align}
and check that $x=15 \implies y=\frac{7\cdot 15}{7} = 15 \quad\checkmark$