Percentage increased in surface area

The surface area $S$ of any similarly shaped solid is directly proportional to the square of its linear dimension $x$ that is $S \propto x^2$ , $\implies \dfrac {S_1}{S_2} = \dfrac {x_1^2}{x_2^2}$ . Then increase in surface area is given by:

$\begin{aligned} \Delta S & = \frac {S_2 - S_1}{S_1} \\ & = \frac {S_2}{S_1} - 1 \\ & = \frac {{\color{#3D99F6}x_2}^2}{x_1^2} - 1 & \small \color{#3D99F6} \text{For }x_2 = x_1 + \Delta x_1 \\ & = \frac {{\color{#3D99F6}(x_1+\Delta x_1)}^2}{x_1^2} - 1 & \small \color{#3D99F6} \text{For }\Delta x_1 = 0.65 x_1 \\ & = \frac {1.65^2 x_1^2}{x_1^2} - 1 \\ & = 1.65^2 - 1 \\ & = 1.7225 \\ & = \boxed{172.25\%} \end{aligned}$

Note: The solutions work for solid of any shape, a cube, a pyramid, a sphere, and even a stone so long as the expansion/contraction is uniform in all three dimensions so as not to affect the shape of the solid.

Let $x$ be the original edge length, then the new edge length is $x+0.65x=1.65x$ . The surface area of the original cube is $6x^2$ . The surface area of the new cube is $6(1.65x)^2=16.335x^2$ . The % increased in surface area is

$\%~increased~in~surface~area = \dfrac{new~surface~area-original~surface~area}{original~surface~area} \times 100 \%=\dfrac{16.335x^2-6x^2}{6x^2} \times 100 \% = \boxed{172.25 \%}$