If each edge of a cube is increased by 50%,

Let a be the edge of the cube.

Then the surface area is $6 a^{2}=S$ (say)

Now, increased edge $=\left(a+\frac{50}{100} a\right)=\frac{150}{100} a=\frac{3}{2} a$

Then, new surface area $=6\left(\frac{3}{2} a\right)^{2}=6 \times \frac{9}{4} a^{2}=\frac{9}{4} S$

In crease in surface area $=\frac{9}{4} S-S=\frac{5}{4} S$

$\therefore$ Percentage increase in surface area $=\frac{\frac{5}{4} S}{S}=\frac{5}{4} \times 100 \%=125 \%$