Sphere Slices

Each 'slice' exposes two new faces. In order to maximize the additional area, these faces need to pass through the center point of the sphere. Their combined area will therefore be $2\pi$ . The initial surface area of a sphere with radius $1$ is $4\pi$ , so we need to find the smallest integer $k$ where $4\pi+k\times{2\pi}>50$ . Simple trial and error yields $\boxed{6}$ .

Surface area of the ball= $4\pi$

for every additonal slice of radius ${r_i}$ we add surface area of $2\pi{r_i}^2$

Total surface area= $4\pi+2\pi\sum_{i=1}^{n} {r_i}^2 >50$

$\sum_{i=1}^{n} {r_i}^2> \frac{25}{\pi}-2 \sim 5.96$

6 very thin plane cuts near center yields a total surface area of about $16\pi \sim 50.26$ . This knife would help.