To solve this problem, we need to determine the fraction of the original activity that remains after a given time, using the concept of half-life.$\\$The half-life $(t_{1/2})$ of the radioactive nuclide is $100$ hours.$\\$The formula to calculate the remaining fraction of a radioactive substance after time $t$ is:$\\ N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \\$Where:$\\$• $N$ is the remaining quantity of the substance.$\\$• $N_0$ is the initial quantity of the substance.$\\$• $t$ is the time elapsed.$\\$• $t_{1/2}$ is the half-life of the substance.$\\$Given:$\\ t = 150$ hours$\\ t_{1/2} = 100$ hours$\\$Substitute these values into the formula:$\\ N = N_0 \left(\frac{1}{2}\right)^{\frac{150}{100}} \\$Simplify the exponent:$\\ N = N_0 \left(\frac{1}{2}\right)^{1.5} \\$Calculate $\left(\frac{1}{2}\right)^{1.5}: \\ \left(\frac{1}{2}\right)^{1.5} = \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{0.5} \\ = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \\ = \frac{1}{2 \sqrt{2}} \\$Thus, the fraction of the original activity that remains after $150$ hours is $\frac{1}{2 \sqrt{2}}. \\$This corresponds to Option 2.