Surface area

let $\color{#D61F06}\large s$ be the surface area of the original cube and $\color{#D61F06}\large S$ be the surface area of the bigger cube

Since the two cubes are similar, we have

$\color{#D61F06}\large \dfrac{s}{S}=\dfrac{x^2}{(1.4x)^2}=\dfrac{x^2}{1.96x^2}=\dfrac{1}{1.96}$ $\large \implies$ $\large \color{#D61F06} S=1.96s$

Therefore,

$\color{#D61F06}\large \%~increased~in~surface~area=(1.96-1)(100\%)=$ $\boxed{\color{#D61F06}\large 96\%}$

Since we are asked about increase in percentage. We can plug in numbers.

Suppose the side of the small cube is $10$ , then its surface area is $6\times 10^2= 600$ . Each dimension is increased by $40\%\implies$ each new dimension is $10\times 1.4=14$ .

The surface area of the new cube is $6\times 14^2= 1176$ .

Thus, the percent increase is: $\large\frac{difference}{original}\times 100\%= \frac{1176-600}{600}\times 100\%= 96\%$