Answer
$\approx \bf 68\%$
Work Step by Step
The probability of finding the electron at a distance $ r > a_B $ is determined by integrating the radial probability density $ P_r(r) $ over the distance from $ a_B $ to infinity.
$$
\text{Prob}(r > a_B) = \int_{a_B}^{\infty} P_r(r) \, dr $$
where $ P_r(r)=4\pi r^2 |R_{1s}(r)|^2$
$$\text{Prob}(r > a_B) = 4\pi \int_{a_B}^{\infty} r^2 |R_{1s}(r)|^2 \, dr\tag 1
$$
Here, $ R_{1s}(r) $ is the radial wave function for the 1s electron in a hydrogen atom, and $ P_r(r) $ is the radial probability density.
The radial wave function for the 1s state is given by:
$$
R_{1s}(r) = \frac{1}{\sqrt{\pi a_B^3}} e^{-r/a_B}
$$
Substituting this into (1);
$$\text{Prob}(r > a_B) = 4\pi \int_{a_B}^{\infty} r^2\left( \frac{1}{\sqrt{\pi a_B^3}} e^{-r/a_B} \right)^2 \, dr
$$
$$\text{Prob}(r > a_B) = 4 \color{red}{\bf\not} \pi \int_{a_B}^{\infty} r^2 \frac{1}{ \color{red}{\bf\not} \pi a_B^3} e^{-2r/a_B} \, dr
$$
$$\text{Prob}(r > a_B) = \frac{4}{ a_B^3} \int_{a_B}^{\infty} r^2 e^{-2r/a_B} \, dr
$$
To simplify the integral, we change the variable. Let $ u = \frac{2r}{a_B} $, so that $ r = \frac{a_B}{2} u $. The differential $ dr $ becomes $dr = \dfrac{a_B}{2} \, du$
Substituting these into the integral and taking the constants out of the integral.
$$
\text{Prob}(r > a_B) = \frac{4}{a_B^3} \left( \frac{a_B^3}{8} \right) \int_2^{\infty} u^2 e^{-u} \, du
$$
So,
$$
\text{Prob}(r > a_B) = \frac{1}{2} \int_2^{\infty} u^2 e^{-u} \, du
$$
This integral can be computed analytically, using integration by parts or standard tables:
$$
\int u^2 e^{-u} \, du = -(u^2 + 2u + 2) e^{-u}
$$
Evaluating this from $ u = 2 $ to $ u = \infty $:
$$
\text{Prob}(r > a_B) = \frac{1}{2} \left[ -(u^2 + 2u + 2) e^{-u} \right]_2^{\infty}
$$
At $ u = \infty $, the exponential term $ e^{-u} $ goes to zero, so we only need to evaluate the lower limit $ u = 2 $:
$$
\text{Prob}(r > a_B) = \frac{1}{2} \left[ -(4 + 4 + 2) e^{-2} \right]
=5e^{-2}
$$
$$
\text{Prob}(r > a_B) = 0.677 \approx \color{red}{\bf 68\%}
$$
