To find the uncertainty in the velocity of the electron within an atomic nucleus of diameter $10^{-15} \ \text{m}$, we can use Heisenberg's uncertainty principle. The principle is expressed as:

$ \Delta x \cdot m \Delta v \geq \frac{h}{4 \pi} $

Here, $\Delta x$ is the uncertainty in position, $m$ is the mass of the electron, $\Delta v$ is the uncertainty in velocity, and $h$ is Planck's constant.

Given:

$ \Delta x = 10^{-15} \ \text{m} $

$ m = 9.1 \times 10^{-31} \ \text{kg} $

$ h = 6.626 \times 10^{-34} \ \text{Js} $

$ \pi = 3.14 $

We need to find $\Delta v$. Rearrange the uncertainty principle to solve for $\Delta v$:

$ \Delta v \approx \frac{h}{4 \pi m \Delta x} $

Plug in the numbers:

$ \Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}} $

First, calculate the denominator:

$ 4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15} = 114.392 \times 10^{-46} $

Now, perform the division:

$ \Delta v = \frac{6.626 \times 10^{-34}}{114.392 \times 10^{-46}} = \frac{6.626}{114.392} \times 10^{12} $

Perform the division:

$ \Delta v \approx 0.0579 \times 10^{12} \ \text{m/s} = 57.97 \times 10^9 \ \text{m/s} $

So, the uncertainty in the velocity of the electron is:

$ 58 \times 10^{9} \ \text{m/s} \quad \text{(nearest integer)} $